# NCERT Solutions Class 11 Maths Chapter 13

## NCERT Solutions for Class 11 Mathematics Chapter 13 – Limits and Derivatives

Mathematics is the subject which helps students to learn the art of patience and the art of perseverance. It develops the strong thinking abilities of the students. As a result, they become capable of coping with the problems more rationally.

Limits and derivatives are an introduction to calculus. Once students have a sound conceptual understanding of this chapter, they will be able to solve all the other chapters of calculus like differentiation, indefinite and definite integration, applications of derivatives and differential equations with ease. The topics like intuitive ideas of derivatives, limits, algebra of limits, limits of trigonometric functions, derivatives, algebra of derivatives of functions, derivatives of polynomials, and trigonometric functions are covered in the NCERT Solutions for Class 11 Mathematics Chapter 13.

The NCERT Solutions have chapter notes related to the entire chapter making it easier for the students to study during their preparation. It is designed while adhering to the latest CBSE syllabus, ensuring they get access to the latest resources. They can also find questions covered from all the segments of the chapter, helping them prepare everything in detail. Extramarks leaves no stones unturned when it comes to providing the best learning material with unmatchable speed and accuracy for students irrespective of the class and subject.Extramarks, is one of the best  online educational platforms, is trusted by   students and teachers  for its good quality resources. Students can find NCERT textbooks, NCERT Exemplar, NCERT solutions, NCERT revision notes, NCERT-related additional questions, CBSE past year papers and mock tests on the Extramarks’ website.

### Key Topics Covered in NCERT Solutions for Class 11 Mathematics Chapter 13

Calculus is the essential part of Class 11 and Class 12 Mathematics as it is interrelated with all the other chapters of Mathematics. Limits and derivatives help in a better understanding of topics like integration and differentiation. The limits and derivatives will be applied while solving problems of differentiation and integration, thereby aiding in making the calculations easier for the students.

NCERT Solutions for Class 11 Mathematics Chapter 13 covers topics like intuitive ideas of derivatives, limits, algebra of limits, limits of trigonometric functions, derivatives, algebra of derivatives of functions, derivatives of polynomials and trigonometric functions. Students can get the NCERT solutions from the Extramarks’ website. After completing this chapter, students will be able to recall all the conditions of limits and be able to  find their derivatives without much difficulty.

Introduction

Limits and derivatives are the initial chapters of Calculus. The chapter begins with the intuitive idea of the derivatives without actually giving their definition. Furthermore, it gives a naive definition of limits and helps students to learn the algebraic sum of the limits. As you proceed, you will learn about the derivatives and their algebraic sum. You will also get to know about derivatives for polynomials and trigonometric functions.

You can learn about this chapter’s concepts and their applications in the NCERT Solutions for Class 11 Mathematics Chapter 13, available on the Extramarks’ website.

The Intuitive Idea of the Derivatives

Here, you will learn to find the object’s velocity with experiments and a few cases and examples. Let us study each of them one by one:

In the first set of computations,

The Average velocity between t = t 1 and t = t2 equals the distance travelled between t = t1 and t = t2 seconds divided by (t 2 – t 1 ).

Hence,

The average velocity in the first two seconds = (Distance travelled between t2 = 2, t1 = 0) / Time interval (t2 – t1)

In the second set of computations,

The average velocities for various time intervals starting at t = 2 seconds.

The average velocity ‘v’ between t = 2 seconds and t = t2  seconds is

= (Distance travelled between 2 seconds and t2 seconds) / (t2 – 2)

= [(Distance travelled in t2 seconds) – (Distance travelled in 2 seconds)] / (t2 – 2)

In the first set of computations, we calculated average velocities in increasing time intervals ending at t = 2 and then hoped that nothing changed just before t = 2.

In the second set of computations, we calculated the average velocities decreasing in time intervals ending at t = 2 and then hoped that nothing changes after t = 2.

Purely on the physical grounds, these sequences of average velocities must approach a common limit.

Limits

The above discussion clearly states the limiting process. Now, let us learn some more about limits with the help of some observations:

We say  limx→a  f (x) is the expected value of f at x = a given the values of f near x to the left of a. This value is called the left-hand limit of f at a.

We say  limx→a f (x) is the expected value of f at x = a given the values of f near x to the right of a. This value is called the right-hand limit of f (x) at a.

If the right and left-hand limits coincide, we call that common value the limit of f (x) at x = a and denote it by limx→a f (x).

Algebra of Limits

Let us learn about the algebra of limits through a theorem without proofs.

Theorem 1:

Let f and g be two functions such that both limx→a f (x) and limx→a g(x) exist.

Then

(i) Limit of the sum of two functions is the sum of the limits of the functions, i.e.,

limx→a [f (x) + g(x)] = limx→a f (x) +  limx→a g(x)

(ii) Limit of difference between two functions is the difference of the limits of the functions, i.e.,

limx→a [f (x)g(x)] = limx→a f (x) limx→a g(x)

(iii) Limit of the product of two functions is the product of the limits of the functions, i.e.,

limx→a [f (x). g(x)] = limx→a f (x).  limx→a g(x)

(iv) Limit of the quotient of two functions is the quotient of the limits of the functions (whenever the denominator is non-zero), i.e.,

limx→a [f (x) / g(x)] =  limx→a f (x) / limx→ag(x)

Limits of polynomials and relational functions

Let us learn about the limits of polynomials through a theorem and draw observations.

The function f is said to be a polynomial function of degree n f(x) = a0 + a1 x + a2 x2 +. . . + an. x n , where a is a real number such that an ≠ 0 for some natural number n. We know that  limx→a x = a.

Hence,

Limx→ax2 = (x-x) = limx→a x . limx→a x = a.a = a2

Theorem 2:

For any positive integer n,

Limx→a(xn-an)/(x-x) = nan-1

Limits of Trigonometric Functions

You can get a clear-cut understanding of the limits of trigonometric functions with the theorems listed below:

Theorem 3:

Let f and g be two real-valued functions with the same domain such that f (x)g( x) for all x in the domain of definition, For some a, if both  Limx→a f(x) and  Limx→a g(x) exist, then  Limx→a f(x) Limx→ag(x).

Theorem 4 (Sandwich theorem):

Let f, g and h be real functions such that f(x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number a, if  Limx→af(x) = l =  Limx→ah(x), then  Limx→ag(x) = l

Theorem 5 :

The following are two important limits.

(i) Limx→0(sin x)/x = 1

(ii)Limx→0(1 – cos x)/x = 0

Derivatives:

Derivatives are the rate of change of a function with respect to the variable. It is applied on all the fundamental units and is quite beneficial in solving the problems of calculus and differential equations.

Definition 1:

Suppose f is a real-valued function and a is a point in its domain of definition. The derivative of f at a is defined by

Limh→0 [f(a+h) − f(a)] / h

Provided this limit exists. Derivative of f (x) at a is denoted by f′(a)

The detailed information about the derivatives is provided in the NCERT Solutions for Class 11 Mathematics Chapter 13, available on the Extramarks’ website.

Algebra of derivatives of functions:

The Algebra of derivatives of functions can be concluded with the help of the following theorem:

Theorem 5:

Let f and g be two functions such that their derivatives are defined in a common domain.

Then,

(i) Derivative of the sum of two functions is the sum of the derivatives of the functions

d/dx.[ f(x) + g(x) ] = d/dx.f(x) + d/dx.g(x)

(ii) Derivative of difference of two functions is the difference of the derivatives of the functions

d/dx.[ f(x) – g(x) ] = d/dx.f(x) – d/dx.g(x)

(iii) Derivative of the product of two functions is given by the following product rule

d/dx.[ f(x). g(x) ] = d/dx.f(x).g(x) + f(x).d/dx.g(x)

(iv) Derivative of the quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero)

d/dx.[ f(x) / g(x) ] = [ d/dx.f(x).g(x) –  f(x).d/dx.g(x) ] / (g(x))2

You can find more information on this topic in the NCERT Solutions for Class 11 Mathematics Chapter 13 available on the Extramarks’ website.

Theorem 6:

Derivative of f(x) = xn is nxn-1 for any positive integer n.

Derivatives of polynomials and trigonometric functions

You can find the derivatives of polynomials and trigonometric functions using the following theorem:

Theorem 7:

Let f(x) = an xn + an-1 xn-1 +…..+ a1 x + a0 be a polynomial function, where ai s are all real numbers and an ≠ 0. Then, the derivative function is given by

df(x) / d(X) = nan xn-1 + (n-1)an-1 xn-2 +….+ 2a2 x + a1

### NCERT Solutions for Class 11 Mathematics Chapter 13 Exercise &  Solutions

You can test your understanding by solving the questions in the NCERT textbook and look for its detailed and in-depth solutions and methodologies in the NCERT Solutions for Class 11 Mathematics Chapter 13. In this way, you can prepare yourselves and be confident for your upcoming examinations. The right solutions and answers provided in it will help you rectify your mistakes and shortcomings. As a result, you will turn into a smart learner. You can look for multiple ways to solve a question and choose the best solution to tackle those tricky and difficult questions in the NCERT solutions.

You can avail of NCERT Solutions for Class 11 Mathematics Chapter 13 from the Extramarks’ website. Click on the  links given below  to view exercise-specific questions and solutions for NCERT Solutions for Class 11 Mathematics Chapter 13:

• Chapter 13 Class 11 Mathematics: Exercise13.1
• Chapter 13 Class 11 Mathematics: Exercise13.2
• Chapter 13 Class 11 Mathematics: Miscellaneous Exercise

Along with Class 11 Mathematics solutions, you can explore NCERT solutions on our Extramarks’ website for all primary and secondary classes

• NCERT Solutions Class 1
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### NCERT Exemplar Class 11 Mathematics

NCERT Exemplar Class 11 Mathematics book has a collection of all NCERT-related questions. Students can find basic to advanced level questions in it. This makes them capable of solving different types of questions. As a result, they will acquire more analytical and critical thinking skills.

The book is specially designed by subject matter experts and gives insights into all the topics from the NCERT Class 11 Mathematics textbook. The benefits provided in this book will overall make a difference not only in-class assignments, tests as well as in the competitive exams later. The simple reason is CBSE itself prescribes NCERT books for board exams and is the best option for competitive exams as well. Thus, they will ensure that even the minutest doubt is resolved and the students develop an interest in learning and mastering the topic with ease.

It helps in laying the foundation for all the basic as well as advanced concepts in the manner required for different competitive examinations. Thus, help to boost the confidence level of the students. They can access  NCERT Exemplar Class 11 Mathematics easily from the Extramarks’ website.

### Key Features for NCERT Solutions for Class 11 Mathematics Chapter 13

Practice makes you perfect. The more you practice, the easier it will get. Extramarks provides a repository of resources with solutions for students to step up their learning and be confident. . Hence, NCERT Solutions for Class 11 Mathematics Chapter 13 provides you with a lot of questions to practice. The key features are as follows:

• Students can find questions from the NCERT textbook, NCERT Exemplar, and other reference books in our NCERT Solutions for Class 11 Mathematics Chapter 13. They will be able to grasp the concepts, think and apply these concepts to solve those tricky questions easily after going through these solutions.
• Experienced subject matter experts have provided answers to all the questions after thoroughly checking and verifying them while strictly adhering to the CBSE guidelines.
• After completing the NCERT Solutions for Class 11 Mathematics Chapter 13, students will be able to solve the chapters like differentiation and integration accurately in tests and exams. It encourages the students to master the topic and achieve high grades..

Q.1 Evaluate the following limits in Exercises 1 to 22.

1.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{3}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{\right)}$

2.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{\pi }}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{x}\mathbf{-}\frac{22}{7}\mathbf{\right)}$

3.

$\underset{\mathbf{r}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}{\mathbf{\pi r}}^{\mathbf{2}}$

4.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{4}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\frac{\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}}{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{\right)}$

5.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{-}\mathbf{1}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\frac{{\mathbf{x}}^{\mathbf{10}}\mathbf{+}{\mathbf{x}}^{\mathbf{5}}\mathbf{+}\mathbf{1}}{\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{\right)}$

6.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\frac{{\mathbf{\left(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{\right)}}^{5}\mathbf{-}\mathbf{1}}{x}$

7.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{2}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{10}}{{\mathbf{x}}^{2}\mathbf{-}\mathbf{4}}\mathbf{\right)}$

8.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{3}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\frac{{\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{81}}{\mathbf{2}{\mathbf{x}}^{2}\mathbf{-}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{3}}$

9.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\frac{\mathbf{ax}\mathbf{+}\mathbf{b}}{\mathbf{cx}\mathbf{+}\mathbf{1}}\mathbf{\right)}$

10.

$\underset{\mathbf{z}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\frac{{\mathbf{z}}^{\frac{1}{3}}\mathbf{-}\mathbf{1}}{{\mathbf{z}}^{\frac{1}{6}}\mathbf{-}\mathbf{1}}$

11.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\frac{{\mathbf{ax}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{c}}{{\mathbf{cx}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{a}}\mathbf{\right)}\mathbf{,}\mathbf{\text{\hspace{0.17em}}}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{\ne }\mathbf{0}$

12.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{-}\mathbf{2}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\frac{\frac{1}{x}\mathbf{+}\frac{\mathbf{1}}{2}}{\mathbf{x}\mathbf{+}\mathbf{2}}$

13.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\frac{\mathrm{sinax}}{\mathrm{bx}}$

14.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\frac{\mathbf{sin}\mathbf{}\mathbf{a}\mathbf{}\mathbf{x}}{\mathbf{sin}\mathbf{}\mathbf{b}\mathbf{}\mathbf{x}}\mathbf{,}\mathbf{\text{\hspace{0.17em}}}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\ne }\mathbf{0}$

15.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{\pi }}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\frac{\mathbf{sin}\mathbf{\left(}\mathbf{\pi }\mathbf{-}\mathbf{x}\mathbf{\right)}}{\mathbf{\pi }\mathbf{\left(}\mathbf{\pi }\mathbf{-}\mathbf{x}\mathbf{\right)}}$

16.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\frac{\mathrm{cos}x}{\mathbf{\left(}\mathbf{\pi }\mathbf{-}\mathbf{x}\mathbf{\right)}}$

17.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\frac{\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{1}}{\mathbf{cosx}\mathbf{-}\mathbf{1}}$

18.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\frac{\mathbf{ax}\mathbf{+}\mathbf{x}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}}{\mathbf{b}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{x}}$

19.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\mathbf{x}\mathbf{}\mathbf{sec}\mathbf{}\mathbf{x}$

20.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\frac{\mathbf{sin}\mathbf{}\mathbf{ax}\mathbf{+}\mathbf{bx}}{\mathbf{ax}\mathbf{+}\mathbf{sin}\mathbf{}\mathbf{bx}}\mathbf{,}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{,}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{\ne }\mathbf{0}$

21.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{cosec}\mathbf{}\mathbf{x}\mathbf{-}\mathbf{cot}\mathbf{\text{\hspace{0.17em}}}\mathbf{x}\mathbf{\right)}$

22.

$\underset{\mathbf{x}\mathbf{\to }\frac{\mathbf{\pi }}{2}}{\mathbf{lim}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\frac{\mathbf{tan}\mathbf{\text{}}\mathbf{2}\mathbf{x}}{\mathbf{x}\mathbf{-}\frac{\mathbf{\pi }}{2}}$

Ans

1.

$\begin{array}{l}\underset{\mathrm{x}\to 3}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\mathrm{x}+3\right)=3+3\\ =6\end{array}$

2.

$\underset{\mathrm{x}\to \mathrm{\pi }}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\mathrm{x}-\frac{22}{7}\right)=\mathrm{\pi }-\frac{22}{7}$

3.

$\begin{array}{l}\underset{\mathrm{r}\to 1}{\mathrm{lim}}\text{\hspace{0.17em}}{\mathrm{\pi r}}^{2}=\mathrm{\pi }{\left(1\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }\end{array}$

4.

$\begin{array}{l}\underset{\mathrm{x}\to 4}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{4\mathrm{x}+3}{\mathrm{x}-2}\right)=\frac{4\left(4\right)+3}{4-2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{19}{2}\end{array}$

5.

$\begin{array}{l}\underset{\mathrm{x}\to -1}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{{\mathrm{x}}^{10}+{\mathrm{x}}^{5}+1}{\mathrm{x}-1}\right)=\frac{{\left(-1\right)}^{10}+{\left(-1\right)}^{5}+1}{\left(-1\right)-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1-1+1}{-1-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\end{array}$

6.

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{{\left(\mathrm{x}+1\right)}^{5}-1}{\mathrm{x}}\\ \mathrm{Let}\text{x}+1=\text{y}⇒\mathrm{x}=\mathrm{y}-1\\ \mathrm{Then},\\ \underset{\mathrm{y}\to 1}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{{\mathrm{y}}^{5}-1}{\mathrm{y}-1}=\underset{\mathrm{y}\to 1}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{{\mathrm{y}}^{5}-{1}^{5}}{\mathrm{y}-1}\\ =5{\left(1\right)}^{5-1}\left[âˆµ\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\frac{{\mathrm{x}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}-\mathrm{a}}={\mathrm{na}}^{\mathrm{n}-1}\right]\end{array}$ $\begin{array}{l}\text{}\text{}\end{array}$ $\begin{array}{l}=5\\ \therefore \underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{{\left(\mathrm{x}+1\right)}^{5}-1}{\mathrm{x}}=5.\end{array}$

7.

$\begin{array}{l}\underset{\mathrm{x}\to 2}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{3{\mathrm{x}}^{2}-\mathrm{x}-10}{{\mathrm{x}}^{2}-4}\right)=\underset{\mathrm{x}\to 2}{\mathrm{lim}}\text{\hspace{0.17em}}\left\{\frac{3{\mathrm{x}}^{2}-6\mathrm{x}+5\mathrm{x}-10}{\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 2}{\mathrm{lim}}\text{\hspace{0.17em}}\left\{\frac{3\mathrm{x}\left(\mathrm{x}-2\right)+5\left(\mathrm{x}-2\right)}{\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 2}{\mathrm{lim}}\text{\hspace{0.17em}}\left\{\frac{\left(\mathrm{x}-2\right)\left(3\mathrm{x}+5\right)}{\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 2}{\mathrm{lim}}\text{\hspace{0.17em}}\left\{\frac{\left(3\mathrm{x}+5\right)}{\left(\mathrm{x}+2\right)}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\left(2\right)+5}{2+2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{11}{4}\end{array}$

8.

$\begin{array}{l}\underset{\mathrm{x}\to 3}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{{\mathrm{x}}^{4}-81}{2{\mathrm{x}}^{2}-5\mathrm{x}-3}=\underset{\mathrm{x}\to 3}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{{\left({\mathrm{x}}^{2}\right)}^{2}-{9}^{2}}{2{\mathrm{x}}^{2}-6\mathrm{x}+\mathrm{x}-3}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 3}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\left({\mathrm{x}}^{2}-9\right)\left({\mathrm{x}}^{2}+9\right)}{2\mathrm{x}\left(\mathrm{x}-3\right)+1\left(\mathrm{x}-3\right)}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\underset{\mathrm{x}\to 3}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\left(\mathrm{x}-3\right)\left(\mathrm{x}+3\right)\left({\mathrm{x}}^{2}+9\right)}{\left(\mathrm{x}-3\right)\left(2\mathrm{x}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 3}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\left(\mathrm{x}+3\right)\left({\mathrm{x}}^{2}+9\right)}{\left(2\mathrm{x}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 3}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\left(3+3\right)\left({3}^{2}+9\right)}{\left(2×3+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{6×18}{7}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{108}{7}\end{array}$

9.

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+1}\right)=\frac{\mathrm{a}\left(0\right)+\mathrm{b}}{\mathrm{c}\left(0\right)+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{b}}{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{b}\end{array}$

10.

$\begin{array}{l}\underset{\mathrm{z}\to 1}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{z}}^{\frac{1}{3}}-1}{{\mathrm{z}}^{\frac{1}{6}}-1}=\underset{\mathrm{z}\to 1}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{{\left({\mathrm{z}}^{\frac{1}{6}}\right)}^{2}-1}{{\mathrm{z}}^{\frac{1}{6}}-1}\\ \text{}=2\left({1}^{2-1}\right)\left[âˆµ\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\frac{{\mathrm{x}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}-\mathrm{a}}={\mathrm{na}}^{\mathrm{n}-1}\right]\\ \text{}=2\end{array}$

11.

$\begin{array}{l}\underset{\mathrm{x}\to 1}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{{\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}}{{\mathrm{cx}}^{2}+\mathrm{bx}+\mathrm{a}}\right)=\frac{\mathrm{a}{\left(1\right)}^{2}+\mathrm{b}\left(1\right)+\mathrm{c}}{\mathrm{c}{\left(1\right)}^{2}+\mathrm{b}\left(1\right)+\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{c}+\mathrm{b}+\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1\end{array}$

12.

$\begin{array}{l}\underset{\mathrm{x}\to -2}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\frac{1}{\mathrm{x}}+\frac{1}{2}}{\mathrm{x}+2}=\underset{\mathrm{x}\to -2}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\frac{2+\mathrm{x}}{2\mathrm{x}}}{\mathrm{x}+2}\\ =\underset{\mathrm{x}\to -2}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\mathrm{x}+2}{2\mathrm{x}\left(\mathrm{x}+2\right)}\\ \underset{\mathrm{x}\to -2}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\frac{1}{\mathrm{x}}+\frac{1}{2}}{\mathrm{x}+2}=\text{\hspace{0.17em}}\frac{1}{2\left(-2\right)}\\ =-\frac{1}{4}\end{array}$

13.

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\mathrm{sin}\mathrm{ax}}{\mathrm{bx}}=\frac{1}{\mathrm{b}}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\frac{\mathrm{sin}\mathrm{ax}}{\mathrm{ax}}×\mathrm{a}\\ =\frac{\mathrm{a}}{\mathrm{b}}\underset{\mathrm{ax}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{\mathrm{sin}\mathrm{ax}}{\mathrm{ax}}\right)\\ =\frac{\mathrm{a}}{\mathrm{b}}×1\left[âˆµ\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\mathrm{x}}{\mathrm{x}}=1\right]\\ =\frac{\mathrm{a}}{\mathrm{b}}\end{array}$

14.

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{sinax}}{\mathrm{sinbx}}=\text{\hspace{0.17em}}\frac{\mathrm{a}\underset{\mathrm{ax}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{\mathrm{sinax}}{\mathrm{ax}}\right)}{\mathrm{b}\underset{\mathrm{bx}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{\mathrm{sinbx}}{\mathrm{bx}}\right)}\\ =\frac{\mathrm{a}×1}{\mathrm{b}×1}\left[âˆµ\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{sinx}}{\mathrm{x}}=1\right]\\ =\frac{\mathrm{a}}{\mathrm{b}}\end{array}$

15.

$\begin{array}{l}\underset{\mathrm{x}\to \mathrm{\pi }}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{sin}\left(\mathrm{\pi }-\mathrm{x}\right)}{\mathrm{\pi }\left(\mathrm{\pi }-\mathrm{x}\right)}=\frac{1}{\mathrm{\pi }}\left\{\underset{\mathrm{\pi }-\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{sin}\left(\mathrm{\pi }-\mathrm{x}\right)}{\left(\mathrm{\pi }-\mathrm{x}\right)}\right\}\\ =\frac{1}{\mathrm{\pi }}\left(1\right)\\ =\frac{1}{\mathrm{\pi }}\end{array}$

16.

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{cosx}}{\left(\mathrm{\pi }-\mathrm{x}\right)}=\frac{\mathrm{cos}0}{\left(\mathrm{\pi }-0\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{\pi }}\left[âˆµ\mathrm{cos}0=1\right]\end{array}$

17.

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{cos}2\mathrm{x}-1}{\mathrm{cosx}-1}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1-2{\mathrm{sin}}^{2}\mathrm{x}-1}{1-2{\mathrm{sin}}^{2}\left(\frac{\mathrm{x}}{2}\right)-1}\\ =\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{-2{\mathrm{sin}}^{2}\mathrm{x}}{-2{\mathrm{sin}}^{2}\left(\frac{\mathrm{x}}{2}\right)}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\end{array}$ $\begin{array}{l}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\left(\frac{{\mathrm{sin}}^{2}\mathrm{x}}{{\mathrm{x}}^{2}}\right)}{\left\{\frac{{\mathrm{sin}}^{2}\left(\frac{\mathrm{x}}{2}\right)}{4{\left(\frac{\mathrm{x}}{2}\right)}^{2}}\right\}}\\ =4\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{{\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)}^{2}}{{\left\{\frac{\mathrm{sin}\left(\frac{\mathrm{x}}{2}\right)}{\left(\frac{\mathrm{x}}{2}\right)}\right\}}^{2}}\\ =4\text{\hspace{0.17em}}\frac{{\left(1\right)}^{2}}{{\left(1\right)}^{2}}\\ \underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{cos}2\mathrm{x}-1}{\mathrm{cosx}-1}=4\end{array}$

18.

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{ax}+\mathrm{x}\mathrm{cos}\mathrm{x}}{\mathrm{b}\mathrm{sin}\mathrm{x}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}\left(\mathrm{a}+\mathrm{cosx}\right)}{\mathrm{b}\mathrm{sin}\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\mathrm{a}+\mathrm{cos}\mathrm{x}\right)}{\mathrm{b}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{\mathrm{sin}\mathrm{x}}{\mathrm{x}}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\left(\mathrm{a}+\mathrm{cos}0\right)}{\mathrm{b}\text{\hspace{0.17em}}\left(1\right)}\\ \underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{ax}+\mathrm{x}\mathrm{cos}\mathrm{x}}{\mathrm{b}\mathrm{sin}\mathrm{x}}=\frac{\mathrm{a}+1}{\mathrm{b}}\end{array}$

19.

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\mathrm{x}\mathrm{sec}\mathrm{x}=0.\mathrm{sec}0\\ \text{\hspace{0.17em}}=0×1\\ \text{\hspace{0.17em}}=0\end{array}$

20.

$\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{sin}\text{}\mathrm{ax}+\mathrm{bx}}{\mathrm{ax}+\mathrm{sin}\mathrm{bx}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}\left(\frac{\mathrm{a}\mathrm{sin}\mathrm{ax}}{\mathrm{ax}}+\mathrm{b}\right)}{\mathrm{x}\left(\mathrm{a}+\frac{\mathrm{b}\mathrm{sin}\mathrm{bx}}{\mathrm{bx}}\right)}$ $\begin{array}{l}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\left(\frac{\mathrm{a}\mathrm{sin}\mathrm{ax}}{\mathrm{ax}}+\mathrm{b}\right)}{\left(\mathrm{a}+\frac{\mathrm{b}\mathrm{sin}\mathrm{bx}}{\mathrm{bx}}\right)}\\ =\text{\hspace{0.17em}}\frac{\left(\mathrm{a}×1+\mathrm{b}\right)}{\left(\mathrm{a}+\mathrm{b}×1\right)}\\ =\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}+\mathrm{b}}\\ \underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{sin}\text{}\mathrm{ax}+\mathrm{bx}}{\mathrm{ax}+\mathrm{sin}\mathrm{bx}}=1\end{array}$

21.

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\mathrm{cosec}\mathrm{x}-\mathrm{cot}\text{\hspace{0.17em}}\mathrm{x}\right)=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{1}{\mathrm{sinx}}-\frac{\mathrm{cos}\text{\hspace{0.17em}}\mathrm{x}}{\mathrm{sinx}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\mathrm{x}}{\mathrm{sinx}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(\frac{2{\mathrm{sin}}^{2}\frac{\mathrm{x}}{2}}{2\mathrm{sin}\frac{\mathrm{x}}{2}\mathrm{cos}\frac{\mathrm{x}}{2}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\frac{\mathrm{x}}{2}}{\mathrm{cos}\frac{\mathrm{x}}{2}}\right)\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(\mathrm{tan}\frac{\mathrm{x}}{2}\right)\\ =\mathrm{tan}0\\ \underset{\mathrm{x}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}}\left(\mathrm{cosec}\mathrm{x}-\mathrm{cot}\text{\hspace{0.17em}}\mathrm{x}\right)=0\end{array}$

22.

$\begin{array}{l}\mathrm{Given}:\text{\hspace{0.17em}\hspace{0.17em}}\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{tan}\text{}2\mathrm{x}}{\mathrm{x}-\frac{\mathrm{\pi }}{2}}\\ \mathrm{Let}\text{x}-\frac{\mathrm{\pi }}{2}=\mathrm{y}\text{}⇒\mathrm{x}=\frac{\mathrm{\pi }}{2}+\mathrm{y}\\ \mathrm{If}\text{x}\to \frac{\mathrm{\pi }}{2}\text{then y}\to 0\\ \therefore \underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{tan}\text{}2\mathrm{x}}{\mathrm{x}-\frac{\mathrm{\pi }}{2}}=\underset{\mathrm{y}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{tan}2\left(\frac{\mathrm{\pi }}{2}+\mathrm{y}\right)}{\frac{\mathrm{\pi }}{2}+\mathrm{y}-\frac{\mathrm{\pi }}{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{y}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{tan}\left(\mathrm{\pi }+2\mathrm{y}\right)}{\mathrm{y}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{y}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{tan}\text{}2\mathrm{y}}{\mathrm{y}}\text{}\left[âˆµ\mathrm{tan}\left(\mathrm{\pi }+\mathrm{x}\right)=\mathrm{tan}\mathrm{x}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\underset{\mathrm{y}\to 0}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{tan}\text{}2\mathrm{y}}{2\mathrm{y}}\end{array}$ $\begin{array}{l}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=2×1\left[\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{tan}\mathrm{x}}{\mathrm{x}}=1\right]\\ \underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{tan}\text{}2\mathrm{x}}{\mathrm{x}-\frac{\mathrm{\pi }}{2}}=2\end{array}$

Q.2

$\mathbf{\text{Find}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\text{and}}\underset{\mathbf{x}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{\text{wheref}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\left\{\begin{array}{l}2x+3,x\le 0\\ 3\left(x+1\right),x>0\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{given}\text{}\mathrm{function}\text{}\mathrm{is}:\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}2\mathrm{x}+3,\mathrm{x}\le 0\\ 3\left(\mathrm{x}+1\right),\mathrm{x}>0\end{array}\\ \mathrm{At}\text{x}=\text{0,}\\ \mathrm{L}.\mathrm{H}.\mathrm{L}.=\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(2\mathrm{x}+3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=2\left(0\right)+3\\ \text{\hspace{0.17em}\hspace{0.17em}}=3\\ \mathrm{R}.\mathrm{H}.\mathrm{L}.=\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}3\left(\mathrm{x}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=3\left(0+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=3\\ \therefore \underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=3\\ \mathrm{At}\text{\hspace{0.17em}}\mathrm{x}=1,\\ \mathrm{L}.\mathrm{H}.\mathrm{L}.=\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 1}{\mathrm{lim}}3\left(\mathrm{x}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=3\left(1+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=6\\ \mathrm{R}.\mathrm{H}.\mathrm{L}.=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 1}{\mathrm{lim}}3\left(\mathrm{x}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=3\left(1+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=6\\ \therefore \underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=6\end{array}$

Q.3

$\mathbf{Find}\phantom{\rule{0ex}{0ex}}\mathbf{}\underset{\mathbf{x}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{where}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{=}\phantom{\rule{0ex}{0ex}}\mathbf{\left\{}\begin{array}{l}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{,}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{\le }\mathbf{1}\\ \mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{,}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{>}\mathbf{1}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{given}\text{}\mathrm{function}\text{}\mathrm{is};\text{f}\left(\text{x}\right)\text{}=\text{}\left\{\begin{array}{l}{\text{x}}^{2}-1,\text{x}\le 1\\ -{\text{x}}^{2}-1,\text{x}>1\end{array}\\ \text{}\mathrm{At}\text{x}=1,\\ \text{L}.\text{H}.\text{L}.=\underset{\text{x}\to {1}^{-}}{\mathrm{lim}}\text{f}\left(\text{x}\right)\\ \text{}=\underset{\text{x}\to 1}{\text{}\mathrm{im}}\left({\text{x}}^{2}-1\right)\\ \text{}=\text{}{\left(1\right)}^{2}-1\\ \text{}=\text{}0\\ \text{R}.\text{H}.\text{L}.=\text{}\underset{\text{x}\to {1}^{+}}{\mathrm{lim}}\text{f}\left(\text{x}\right)\\ \text{}=\text{}\underset{\text{x}\to 1}{\mathrm{lim}}\left(-{\text{x}}^{2}-1\right)\\ \text{}=\text{}-{\left(1\right)}^{2}-1\\ \text{}=\text{}-2\\ \therefore \underset{\text{x}\to {1}^{-}}{\mathrm{lim}}\text{f}\left(\text{x}\right)\underset{\text{x}\to {1}^{+}}{\mathrm{lim}}\text{f}\left(\text{x}\right)\\ \mathrm{Thus},\text{}\underset{\text{x}\to 1}{\mathrm{lim}}\text{f}\left(\text{x}\right)\text{}\mathrm{does}\text{}\mathrm{not}\text{}\mathrm{exist}.\end{array}$

Q.4

$\mathbf{Find}\mathbf{\text{}}\underset{\mathbf{\text{x}}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{f}}\left(\text{x}\right)\mathbf{,}\mathbf{\text{}}\mathbf{}\mathbf{where}\mathbf{\text{}}\mathbf{}\mathbf{\text{f}}\left(\text{x}\right)\mathbf{\text{}}\mathbf{=}\mathbf{\text{}}\mathbf{\left\{}\begin{array}{lll}\frac{\mathbf{\text{x}}}{|\text{x}|}\mathbf{,}& & \mathbf{\text{x}}\mathbf{\ne }\mathbf{0}\\ \mathbf{0}\mathbf{,}& & \mathbf{\text{x}}\mathbf{=}\mathbf{0}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{}\mathrm{given}\text{}\mathrm{function}\text{}\mathrm{is},\\ \text{f}\left(\text{x}\right)=\left\{\begin{array}{lll}\frac{\text{x}}{\left|\text{x}\right|},& & \text{x}\ne 0\\ 0,& & \text{x}=0\end{array}\\ \text{Atx}=0,\\ \text{L}.\text{H}.\text{L}.\text{}=\underset{\text{x}\to {0}^{-}}{\text{lim}}\text{f}\left(\text{x}\right)\\ \text{}=\underset{\text{x}\to {0}^{-}}{\text{lim}}\left(\frac{\text{x}}{\left|\text{x}\right|}\right)\\ \text{}=\underset{\text{x}\to 0}{\text{lim}}\left(\frac{\text{x}}{-\text{x}}\right)\text{}\left[\text{whenx}<0,\text{}\left|\text{x}\right|\text{}=\text{}-\text{x}\right]\\ \text{}=-1\\ \text{R}.\text{H}.\text{L}.=\underset{\text{x}\to {0}^{+}}{\text{lim}}\text{f}\left(\text{x}\right)\\ \text{}=\underset{\text{x}\to {0}^{+}}{\text{lim}}\left(\frac{\text{x}}{\left|\text{x}\right|}\right)\\ \text{}=\underset{\text{x}\to 0}{\text{lim}}\left(\frac{\text{x}}{\text{x}}\right)\text{}\left[\text{Whenx}>0,\text{}\left|\text{x}\right|\text{}=\text{x}\right]\\ \text{}=1\\ \text{Since},\text{}\underset{\text{x}\to {0}^{-}}{\text{lim}}\text{f}\left(\text{x}\right)\text{}\underset{\text{x}\to {0}^{+}}{\text{lim}}\text{f}\left(\text{x}\right)\\ \text{So,}\underset{\text{x}\to 0}{\text{lim}}\text{f}\left(\text{x}\right)\text{doesnotexist}.\end{array}$

Q.5

$\begin{array}{l}\mathbf{Suppose}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\begin{array}{l}\mathbf{a}\mathbf{+}\mathbf{bx}\mathbf{,}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{<}\mathbf{1}\\ \mathbf{4}\mathbf{,}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\mathbf{1}\\ \mathbf{b}\mathbf{-}\mathbf{ax}\mathbf{,}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{>}\mathbf{1}\end{array}\\ \mathbf{and}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{if}\phantom{\rule{0ex}{0ex}}\underset{\mathbf{x}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{f}\phantom{\rule{0ex}{0ex}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{=}\phantom{\rule{0ex}{0ex}}\mathbf{f}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{what}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{are}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{possible}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{values}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{of}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{a}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{and}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{b}\mathbf{?}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{given function is:}\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{a}+\mathrm{bx},\mathrm{x}<1\\ 4,\mathrm{x}=1\\ \mathrm{b}-\mathrm{ax},\mathrm{x}>1\end{array}\\ \mathrm{For}\text{x}=\text{1,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}L.H.L.}=\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 1}{\mathrm{lim}}\left(\mathrm{a}+\mathrm{bx}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{a}+\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}R.H.L.}=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 1}{\mathrm{lim}}\left(\mathrm{b}-\mathrm{ax}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{b}-\mathrm{a}\\ \mathrm{and}\text{\hspace{0.17em}}\mathrm{f}\left(1\right)=4\\ \mathrm{According}\text{to given condition,}\\ \underset{\mathrm{x}\to 1}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(1\right)\\ ⇒\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(1\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\mathrm{b}\text{\hspace{0.17em}}=\mathrm{b}-\mathrm{a}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\mathrm{b}\text{\hspace{0.17em}}=4\text{and}\mathrm{b}-\mathrm{a}=4\\ \mathrm{Solving}\text{both equations, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}=0\text{and}\mathrm{b}=4\\ \mathrm{Thus},\text{the possible values of a and b are 0 and 4 respectively.}\end{array}$

Q.6

$\mathbf{Find}\mathbf{}\mathbf{\text{}}\underset{\mathbf{x}\mathbf{\to }\mathbf{5}}{\mathbf{lim}}\mathbf{\text{}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{where}\mathbf{}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{|}\mathbf{x}\mathbf{|}\mathbf{-}\mathbf{5}$

Ans

$\begin{array}{l}\mathrm{The}\text{given function is,}\\ \text{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=|\mathrm{x}|-5=\left\{\begin{array}{l}-\mathrm{x}-5,\mathrm{x}<0\\ -5,\mathrm{x}=0\\ \mathrm{x}-5,\mathrm{x}>0\end{array}\\ \mathrm{For}\text{x}=5,\\ \mathrm{L}.\mathrm{H}.\mathrm{L}.=\underset{\mathrm{x}\to {5}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 5}{\mathrm{lim}}\left(\mathrm{x}-5\right)\left[\mathrm{When}\text{x>0,}|\mathrm{x}|=\mathrm{x}\right]\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=5-5\\ \text{\hspace{0.17em}\hspace{0.17em}}=0\\ \mathrm{R}.\mathrm{H}.\mathrm{L}.=\underset{\mathrm{x}\to {5}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 5}{\mathrm{lim}}\left(\mathrm{x}-5\right)\left[\mathrm{When}\text{x>0,}|\mathrm{x}|=\mathrm{x}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=5-5\\ \text{\hspace{0.17em}\hspace{0.17em}}=0\\ \therefore \underset{\mathrm{x}\to {5}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {5}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 5}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=0\\ \mathrm{Thus},\text{}\underset{\mathrm{x}\to 5}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\text{0}\end{array}$

Q.7 Let a1, a2, …, an be fixed real numbers and define a function f(x) = (x – a1) (x – a2)… (x – an).

$\mathbf{What}\mathbf{}\mathbf{is}\underset{\mathbf{x}\mathbf{\to }{\mathbf{a}}_{\mathbf{1}}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{?}\mathbf{}\mathbf{For}\mathbf{}\mathbf{some}\mathbf{}\mathbf{a}\mathbf{\ne }{\mathbf{a}}_{\mathbf{1}}\mathbf{,}\mathbf{\text{\hspace{0.17em}}}{\mathbf{a}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{..}\mathbf{,}{\mathbf{a}}_{\mathbf{n}}\mathbf{,}\mathbf{compute}\mathbf{}\underset{\mathbf{x}\mathbf{\to }\mathbf{a}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{The}\text{given function is:}\\ \text{f}\left(\mathrm{x}\right)=\left(\mathrm{x}-{\mathrm{a}}_{1}\right)\left(\mathrm{x}-{\mathrm{a}}_{2}\right)\dots \left(\mathrm{x}-{\mathrm{a}}_{\mathrm{n}}\right)\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{\mathrm{x}\to {\mathrm{a}}_{1}}{\mathrm{lim}}\text{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}_{1}}{\mathrm{lim}}\left\{\left(\mathrm{x}-{\mathrm{a}}_{1}\right)\left(\mathrm{x}-{\mathrm{a}}_{2}\right)\dots \left(\mathrm{x}-{\mathrm{a}}_{\mathrm{n}}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\underset{\mathrm{x}\to {\mathrm{a}}_{1}}{\mathrm{lim}}\left(\mathrm{x}-{\mathrm{a}}_{1}\right)\right\}\left\{\underset{\mathrm{x}\to {\mathrm{a}}_{1}}{\mathrm{lim}}\left(\mathrm{x}-{\mathrm{a}}_{2}\right)\right\}\dots \left\{\underset{\mathrm{x}\to {\mathrm{a}}_{1}}{\mathrm{lim}}\left(\mathrm{x}-{\mathrm{a}}_{\mathrm{n}}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left({\mathrm{a}}_{1}-{\mathrm{a}}_{1}\right)\left({\mathrm{a}}_{1}-{\mathrm{a}}_{2}\right)\dots \left({\mathrm{a}}_{1}-{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(0\right)\left({\mathrm{a}}_{1}-{\mathrm{a}}_{2}\right)\dots \left({\mathrm{a}}_{1}-{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\end{array}$ $\mathrm{And}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\text{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\left\{\left(\mathrm{x}-{\mathrm{a}}_{1}\right)\left(\mathrm{x}-{\mathrm{a}}_{2}\right)\dots \left(\mathrm{x}-{\mathrm{a}}_{\mathrm{n}}\right)\right\}$ $\begin{array}{l}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\left\{\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\left(\mathrm{x}-{\mathrm{a}}_{1}\right)\right\}\left\{\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\left(\mathrm{x}-{\mathrm{a}}_{2}\right)\right\}\dots \left\{\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\left(\mathrm{x}-{\mathrm{a}}_{\mathrm{n}}\right)\right\}\\ =\left(\mathrm{a}-{\mathrm{a}}_{1}\right)\left(\mathrm{a}-{\mathrm{a}}_{2}\right)\dots \left(\mathrm{a}-{\mathrm{a}}_{\mathrm{n}}\right)\end{array}$

Q.8

$\begin{array}{l}\mathbf{If}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\begin{array}{ll}\mathbf{|}\mathbf{x}\mathbf{|}\mathbf{+}\mathbf{1}\mathbf{,}& \mathbf{x}\mathbf{<}\mathbf{0}\\ \mathbf{0}\mathbf{,}& \mathbf{x}\mathbf{=}\mathbf{0}\\ \mathbf{|}\mathbf{x}\mathbf{|}\mathbf{-}\mathbf{1}\mathbf{,}& \mathbf{x}\mathbf{>}\mathbf{0}\end{array}\\ \mathbf{For}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{what}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{value}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{of}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{a}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{does}\mathbf{}\phantom{\rule{0ex}{0ex}}\underset{\mathbf{x}\mathbf{\to }\mathbf{a}}{\mathbf{lim}}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{exists}\mathbf{?}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{given function is:}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}|\mathrm{x}|+1,\mathrm{x}<0\\ 0,\mathrm{x}=0\\ |\mathrm{x}|-1,\mathrm{x}>0\end{array}\\ \text{For existence of}\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right),\end{array}$ $\begin{array}{l}\text{CaseI:Whena=0}\\ \underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\left(\left|\mathrm{x}\right|+1\right)\\ \text{}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(-\mathrm{x}+1\right)\text{}\left[\mathrm{When}\text{\hspace{0.17em}}\mathrm{x}<0,\text{\hspace{0.17em}}\left|\mathrm{x}\right|=-\mathrm{x}\right]\\ \text{}=0+1\\ \text{}=1\\ \underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\left(\left|\mathrm{x}\right|-1\right)\\ \text{}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(\mathrm{x}-1\right)\text{}\left[\mathrm{When}\text{\hspace{0.17em}}\mathrm{x}>0,\text{\hspace{0.17em}}\left|\mathrm{x}\right|=\mathrm{x}\right]\\ \text{}=0-1\\ \text{}=-1\\ \mathrm{Here},\text{}\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\ne \underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \therefore \underset{\mathrm{x}\to 0}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\text{does not exist.}\\ \mathrm{Case}\text{\hspace{0.17em}}\mathrm{II}\text{}:\text{}\mathrm{When}\text{a}<0\\ \underset{\mathrm{x}\to {\mathrm{a}}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{\mathrm{lim}}\left(\left|\mathrm{x}\right|+1\right)\\ \text{}=\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\left(-\text{\hspace{0.17em}}\mathrm{x}+1\right)\text{}\left[\mathrm{When}\text{\hspace{0.17em}}\mathrm{x}<0,\text{\hspace{0.17em}}\left|\mathrm{x}\right|=-\mathrm{x}\right]\\ \text{}=-\text{\hspace{0.17em}}\mathrm{a}+1\\ \underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\mathrm{lim}}\left(\left|\mathrm{x}\right|+1\right)\\ =\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\left(-\mathrm{x}+1\right)\text{}\left[\mathrm{When}\text{\hspace{0.17em}}\mathrm{a}<\mathrm{x}<0,\text{\hspace{0.17em}}\left|\mathrm{x}\right|=-\mathrm{x}\right]\\ =-\text{\hspace{0.17em}}\mathrm{a}+1\\ \therefore \underset{\mathrm{x}\to {\mathrm{a}}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=-\text{\hspace{0.17em}}\mathrm{a}+1\end{array}$ $\begin{array}{l}\mathrm{Thus},\text{}\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\text{exists at x}=\text{a, where a<0.}\\ \mathrm{Case}\text{\hspace{0.17em}}\mathrm{III}:\text{\hspace{0.17em}}\mathrm{When}\text{a}>0\\ \underset{\mathrm{x}\to {\mathrm{a}}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{\mathrm{lim}}\left(|\mathrm{x}|-1\right)\\ =\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\left(\text{\hspace{0.17em}}\mathrm{x}-1\right)\left[\mathrm{When}\text{\hspace{0.17em}}0<\text{\hspace{0.17em}}\mathrm{x}<\mathrm{a},\text{\hspace{0.17em}}|\mathrm{x}|=\mathrm{x}\right]\\ =\text{\hspace{0.17em}}\mathrm{a}-1\\ \underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\mathrm{lim}}\left(|\mathrm{x}|-1\right)\\ =\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\left(\mathrm{x}-1\right)\left[\mathrm{When}\text{\hspace{0.17em}}\mathrm{a}<\mathrm{x}<0,\text{\hspace{0.17em}}|\mathrm{x}|=\mathrm{x}\right]\\ =\text{\hspace{0.17em}}\mathrm{a}-1\\ \therefore \underset{\mathrm{x}\to {\mathrm{a}}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}\mathrm{a}-1\\ \mathrm{Thus},\text{}\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\text{exists at x}=\text{a, where a>0.}\\ \mathrm{Therefore},\text{}\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\text{\hspace{0.17em}\hspace{0.17em}exists for all a}\ne \text{0.}\end{array}$

Q.9

$\begin{array}{l}\mathrm{If}\text{}\mathrm{the}\text{}\mathrm{function}\text{f}\left(\text{x}\right)\text{}\mathrm{satisfies}\text{}\underset{\text{x}\to 1}{\mathrm{lim}}\text{}\frac{\text{f}\left(\text{x}\right)-2}{{\text{x}}^{2}-1}=\text{π},\\ \mathrm{evaluate}\text{}\underset{\text{x}\to 1}{\mathrm{lim}}\text{f}\left(\text{x}\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{G}\text{iven: \hspace{0.17em}}\underset{\mathrm{x}\to 1}{\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}\right)-2}{{\mathrm{x}}^{2}-1}=\mathrm{\pi }\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{\mathrm{x}\to 1}{\mathrm{lim}}\left\{\mathrm{f}\left(\mathrm{x}\right)-2\right\}=\underset{\mathrm{x}\to 1}{\mathrm{lim}}\mathrm{\pi }\left({\mathrm{x}}^{2}-1\right)\\ ⇒\text{\hspace{0.17em}}\underset{\mathrm{x}\to 1}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)-\underset{\mathrm{x}\to 1}{\mathrm{lim}}2\text{\hspace{0.17em}}=\mathrm{\pi }\left({1}^{2}-1\right)\\ ⇒\text{\hspace{0.17em}}\underset{\mathrm{x}\to 1}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)-\underset{\mathrm{x}\to 1}{\mathrm{lim}}2\text{\hspace{0.17em}}=\mathrm{\pi }\left(0\right)\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{\mathrm{x}\to 1}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\mathrm{lim}}2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{\mathrm{x}\to 1}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=2.\end{array}$

Q.10

$\begin{array}{l}\mathrm{If}\text{f}\left(\text{x}\right)=\left\{\begin{array}{lll}{\text{mx}}^{2}+\text{n},& & \text{x}<0\\ \mathrm{nx}+\text{m},& & 0\le \text{x}\le 1\\ {\text{nx}}^{3}+\text{m},& & \text{x}>0\end{array}.\\ \mathrm{For}\text{}\mathrm{what}\text{}\mathrm{integers}\text{m}\mathrm{and}\text{n}\mathrm{does}\text{}\mathrm{both}\text{}\underset{\text{x}\to 0}{\mathrm{lim}}\text{f}\left(\text{x}\right)\text{}\mathrm{and}\text{}\underset{\text{x}\to 1}{\mathrm{lim}}\text{f}\left(\text{x}\right)\text{}\mathrm{exist}?\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\mathrm{mx}}^{2}+\mathrm{n},\mathrm{x}<0\\ \mathrm{nx}+\mathrm{m},0\le \mathrm{x}\le 1\\ {\mathrm{nx}}^{3}+\mathrm{m},\mathrm{x}>0\end{array}\\ \mathrm{For}\text{x}=\text{0,}\\ \underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left({\mathrm{mx}}^{2}+\mathrm{n}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}\\ \underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left({\mathrm{nx}}^{3}+\mathrm{m}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{m}\\ \mathrm{Since},\text{}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\text{exists. So,}\\ \underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=\mathrm{m}\\ \mathrm{Therefore},\text{}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\text{exists if m}=\text{n.}\\ \mathrm{For}\text{x}=\text{1,}\end{array}$ $\begin{array}{l}\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\mathrm{lim}}\left(\mathrm{nx}+\mathrm{m}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}+\mathrm{m}\\ \underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\mathrm{lim}}\left({\mathrm{nx}}^{3}+\mathrm{m}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}+\mathrm{m}\\ \mathrm{Since},\text{}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\text{exists. So,}\\ \underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \mathrm{Therefore},\text{}\underset{\mathrm{x}\to 1}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\text{exists}\mathrm{for}\text{any value of m and n.}\end{array}$

Q.11 Find the derivative of x2 – 2 at x = 10.

Ans

$\begin{array}{l}\mathrm{We}\text{are given:}\\ \text{\hspace{0.17em}\hspace{0.17em}f}\left(\mathrm{x}\right)={\mathrm{x}}^{2}-2\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(10\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(10+\mathrm{h}\right)-\mathrm{f}\left(10\right)}{\mathrm{h}}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(10\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\left\{{\left(10+\mathrm{h}\right)}^{2}-2\right\}-\left\{{\left(10\right)}^{2}-2\right\}}{\mathrm{h}}\\ \text{\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\left(100+20\mathrm{h}+{\mathrm{h}}^{2}-2\right)-\left(100-2\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{98+20\mathrm{h}+{\mathrm{h}}^{2}-98}{\mathrm{h}}\end{array}$ $\begin{array}{l}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{h}\left(20+\mathrm{h}\right)}{\mathrm{h}}\\ \mathrm{f}‘\left(10\right)=20\\ \mathrm{Thus},{\text{the derivative of the function x}}^{\text{2}}-\text{2 at x}=\text{10 is 20.}\end{array}$

Q.12 Find the derivative of 99x at x = l00.

Ans

$\begin{array}{l}\text{Wearegiven:f}\left(\text{x}\right)\text{=99x}\\ \text{Since,}\mathrm{f}‘\left(100\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(100+\mathrm{h}\right)-\mathrm{f}\left(100\right)}{\mathrm{h}}\\ \therefore \mathrm{f}‘\left(100\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{99\left(100+\mathrm{h}\right)-99\left(100\right)}{\mathrm{h}}\\ \text{}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{9900+99\mathrm{h}-9900}{\mathrm{h}}\\ \text{}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{99\mathrm{h}}{\mathrm{h}}\\ \text{}\mathrm{f}‘\left(100\right)=99\\ \text{Thus,thederivativeofthefunction99xatx=100is99.}\end{array}$

Q.13 Find the derivative of x at x = 1.

Ans

$\begin{array}{l}\text{Wearegiven:}\\ \text{f}\left(\mathrm{x}\right)=\mathrm{x}\\ \text{Since,}\mathrm{f}‘\left(1\right)=\underset{}{\underset{\mathrm{h}\to 0}{\mathrm{lim}}}\frac{\mathrm{f}\left(1+\mathrm{h}\right)-\mathrm{f}\left(1\right)}{\mathrm{h}}\\ \therefore \text{}\mathrm{f}‘\left(1\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\left(1+\mathrm{h}\right)-\left(1\right)}{\mathrm{h}}\\ \text{}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{1+\mathrm{h}-1}{\mathrm{h}}\\ \text{}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{h}}{\mathrm{h}}\\ \text{}\mathrm{f}‘\left(1\right)=1\\ \text{Thus,thederivativeofthefunctionxatx=1is1.}\end{array}$

Q.14 Find the derivative of the following functions from first principle.

$\begin{array}{l}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{27}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{\left(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{iii}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{iv}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{x}\mathbf{+}\mathbf{1}}{\mathbf{x}\mathbf{-}\mathbf{1}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\\ \mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{3}-27\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\left\{{\left(\mathrm{x}+\mathrm{h}\right)}^{3}-27\right\}-\left({\mathrm{x}}^{3}-27\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\mathrm{x}}^{3}+3{\mathrm{x}}^{2}\mathrm{h}+3{\mathrm{xh}}^{2}+{\mathrm{h}}^{3}-27-{\mathrm{x}}^{3}+27}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{3{\mathrm{x}}^{2}\mathrm{h}+3{\mathrm{xh}}^{2}+{\mathrm{h}}^{3}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\left(3{\mathrm{x}}^{2}+3\mathrm{xh}+{\mathrm{h}}^{2}\right)\mathrm{h}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3{\mathrm{x}}^{2}+3\mathrm{x}\left(0\right)+{\left(0\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=3{\mathrm{x}}^{2}\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}-3\mathrm{x}+2\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{x}+\mathrm{h}\right)}^{2}-3\left(\mathrm{x}+\mathrm{h}\right)+2-\left({\mathrm{x}}^{2}-3\mathrm{x}+2\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\mathrm{x}}^{2}+2\mathrm{xh}+{\mathrm{h}}^{2}-3\mathrm{x}-3\mathrm{h}+2-{\mathrm{x}}^{2}+3\mathrm{x}-2}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{2\mathrm{xh}+{\mathrm{h}}^{2}-3\mathrm{h}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\left(2\mathrm{x}+\mathrm{h}-3\right)\mathrm{h}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}-3\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^{2}}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\frac{1}{{\left(\mathrm{x}+\mathrm{h}\right)}^{2}}-\frac{1}{{\mathrm{x}}^{2}}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\mathrm{x}}^{2}-{\left(\mathrm{x}+\mathrm{h}\right)}^{2}}{\mathrm{h}{\left(\mathrm{x}+\mathrm{h}\right)}^{2}{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\mathrm{x}}^{2}-{\mathrm{x}}^{2}-2\mathrm{xh}-{\mathrm{h}}^{2}}{\mathrm{h}{\left(\mathrm{x}+\mathrm{h}\right)}^{2}{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{-2\mathrm{xh}-{\mathrm{h}}^{2}}{\mathrm{h}{\left(\mathrm{x}+\mathrm{h}\right)}^{2}{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{-\mathrm{h}\left(2\mathrm{x}+\mathrm{h}\right)}{\mathrm{h}{\left(\mathrm{x}+\mathrm{h}\right)}^{2}{\mathrm{x}}^{2}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-\left(2\mathrm{x}+0\right)}{{\left(\mathrm{x}+0\right)}^{2}{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{2\mathrm{x}}{{\mathrm{x}}^{4}}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\frac{2}{{\mathrm{x}}^{3}}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+1}{\mathrm{x}-1}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\frac{\mathrm{x}+\mathrm{h}+1}{\left(\mathrm{x}+\mathrm{h}-1\right)}-\frac{\mathrm{x}+1}{\left(\mathrm{x}-1\right)}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\left(\mathrm{x}+\mathrm{h}+1\right)\left(\mathrm{x}-1\right)-\left(\mathrm{x}+1\right)\left(\mathrm{x}+\mathrm{h}-1\right)}{\mathrm{h}\left(\mathrm{x}+\mathrm{h}-1\right)\left(\mathrm{x}-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\left({\mathrm{x}}^{2}+\mathrm{xh}+\mathrm{x}-\mathrm{x}-\mathrm{h}-1\right)-\left({\mathrm{x}}^{2}+\mathrm{xh}-\mathrm{x}+\mathrm{x}+\mathrm{h}-1\right)}{\mathrm{h}\left(\mathrm{x}+\mathrm{h}-1\right)\left(\mathrm{x}-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\mathrm{x}}^{2}+\mathrm{xh}+\mathrm{x}-\mathrm{x}-\mathrm{h}-1-{\mathrm{x}}^{2}-\mathrm{xh}+\mathrm{x}-\mathrm{x}-\mathrm{h}+1}{\mathrm{h}\left(\mathrm{x}+\mathrm{h}-1\right)\left(\mathrm{x}-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{-2\mathrm{h}}{\mathrm{h}\left(\mathrm{x}+\mathrm{h}-1\right)\left(\mathrm{x}-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-2}{\left(\mathrm{x}+0-1\right)\left(\mathrm{x}-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\frac{-2}{{\left(\mathrm{x}-1\right)}^{2}}\end{array}$

Q.15

$\mathbf{For}\mathbf{}\mathbf{the}\mathbf{}\mathbf{function}\mathbf{}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{100}}}{\mathbf{100}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{99}}}{\mathbf{99}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{}\mathbf{Prove}\mathbf{}\mathbf{that}\mathbf{}\mathbf{\text{\hspace{0.17em}}}\mathbf{f}\mathbf{‘}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathbf{}\mathbf{f}\mathbf{‘}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{100}}{100}+\frac{{\mathrm{x}}^{99}}{99}+...+\frac{{\mathrm{x}}^{2}}{2}+\mathrm{x}+1\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{x}}^{100}}{100}+\frac{{\mathrm{x}}^{99}}{99}+...+\frac{{\mathrm{x}}^{2}}{2}+\mathrm{x}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{d}}{\mathrm{dx}}\frac{{\mathrm{x}}^{100}}{100}+\frac{\mathrm{d}}{\mathrm{dx}}\frac{{\mathrm{x}}^{99}}{99}+...+\frac{\mathrm{d}}{\mathrm{dx}}\frac{{\mathrm{x}}^{2}}{2}+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}}1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{100{\mathrm{x}}^{99}}{100}+\frac{99{\mathrm{x}}^{98}}{99}+...+\frac{2{\mathrm{x}}^{1}}{2}+1+0\left[\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{\mathrm{n}}={\mathrm{nx}}^{\mathrm{n}-1}\right]\\ \mathrm{f}‘\left(\mathrm{x}\right)={\mathrm{x}}^{99}+{\mathrm{x}}^{98}+...+\mathrm{x}+1\\ \mathrm{Substituting}\text{x}=\text{1, we get}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(1\right)={\left(1\right)}^{99}+{\left(1\right)}^{98}+...+\left(1\right)+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=100\\ \mathrm{Substituting}\text{x}=\text{0, we get}\\ \mathrm{f}‘\left(0\right)={\left(0\right)}^{99}+{\left(0\right)}^{98}+...+\left(0\right)+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\\ 100\mathrm{f}‘\left(0\right)=100\left(1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=100\end{array}$ $\begin{array}{l}\mathrm{Thus},\text{\hspace{0.17em}}\mathrm{f}‘\left(1\right)=100\mathrm{f}‘\left(0\right).\\ \mathrm{Therefore},\text{it is proved.}\end{array}$

Q.16 Find the derivative of xn + axn−1 + a2xn−2 + . . .+ an−1x + an for some fixed real number a.

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)={\text{x}}^{\text{n}}+{{\text{ax}}^{\text{n}}}^{-\text{1}}+{\text{a}}^{\text{2}}{{\text{x}}^{\text{n}}}^{-\text{2}}+.\text{}.\text{}.+{{\text{a}}^{\text{n}}}^{-\text{1}}\text{x}+{\text{a}}^{\text{n}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left({\text{x}}^{\text{n}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left({{\text{ax}}^{\text{n}}}^{-\text{1}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left({\text{a}}^{\text{2}}{{\text{x}}^{\text{n}}}^{-\text{2}}\right)+.\text{}.\text{}.+\frac{\mathrm{d}}{\mathrm{dx}}\left({{\text{a}}^{\text{n}}}^{-\text{1}}\text{x}\right)\\ +\frac{\mathrm{d}}{\mathrm{dx}}{\text{a}}^{\text{n}}\\ ={\mathrm{nx}}^{\mathrm{n}-1}+\mathrm{a}\left(\mathrm{n}-1\right){\mathrm{x}}^{\mathrm{n}-2}+{\mathrm{a}}^{2}\left(\mathrm{n}-2\right){\mathrm{x}}^{\mathrm{n}-3}+...+{\mathrm{a}}^{\mathrm{n}-1}+0\\ ={\mathrm{nx}}^{\mathrm{n}-1}+\mathrm{a}\left(\mathrm{n}-1\right){\mathrm{x}}^{\mathrm{n}-2}+{\mathrm{a}}^{2}\left(\mathrm{n}-2\right){\mathrm{x}}^{\mathrm{n}-3}+...+{\mathrm{a}}^{\mathrm{n}-1}\end{array}$

Q.17

$\mathbf{\text{Find}}\underset{\mathbf{\text{x}}\mathbf{\to }\mathbf{\text{0}}}{\mathbf{\text{lim}}}\mathbf{\text{f}}\left(\text{x}\right)\mathbf{\text{,wheref}}\left(\text{x}\right)\mathbf{\text{}}\mathbf{=}\mathbf{\text{}}\left\{\begin{array}{c}\frac{x}{|x|},x\ne 0\\ 0,x=0\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{given function is,}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\mathrm{x}}{|\mathrm{x}|},\mathrm{x}\ne 0\\ 0,\mathrm{x}=0\end{array}\\ \mathrm{At}\text{x}=0,\\ \mathrm{L}.\mathrm{H}.\mathrm{L}.=\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\left(\frac{\mathrm{x}}{|\mathrm{x}|}\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(\frac{\mathrm{x}}{-\mathrm{x}}\right)\left[\mathrm{when}\text{\hspace{0.17em}}\mathrm{x}<0,\text{\hspace{0.17em}}|\mathrm{x}|=-\mathrm{x}\right]\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-1\\ \mathrm{R}.\mathrm{H}.\mathrm{L}.=\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\left(\frac{\mathrm{x}}{|\mathrm{x}|}\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(\frac{\mathrm{x}}{\mathrm{x}}\right)\left[\mathrm{when}\text{\hspace{0.17em}}\mathrm{x}<0,\text{\hspace{0.17em}}|\mathrm{x}|=\mathrm{x}\right]\end{array}$ $\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1\\ \mathrm{Since},\text{\hspace{0.17em}}\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\ne \underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\\ \mathrm{So},\text{}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\mathrm{f}\left(\mathrm{x}\right)\text{does not exist.}\end{array}$

Q.18

$\begin{array}{l}\mathbf{\text{For some constants a and b, find the derivative of}}\\ \left(\text{i}\right)\mathbf{\text{}}\left(\text{x}-\text{a}\right)\left(\text{x}-\text{b}\right)\mathbf{\text{}}\left(\text{ii}\right)\mathbf{\text{}}{\left({\text{ax}}^{2}+\text{b}\right)}^{\mathbf{2}}\mathbf{\text{}}\left(\text{iii}\right)\mathbf{\text{}}\frac{\mathbf{\text{x}}\mathbf{-}\mathbf{\text{a}}}{\mathbf{\text{x}}\mathbf{-}\mathbf{\text{b}}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}-\mathrm{a}\right)\left(\mathrm{x}-\mathrm{b}\right)\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left(\mathrm{x}-\mathrm{a}\right)\left(\mathrm{x}-\mathrm{b}\right)\right\}\\ =\left(\mathrm{x}-\mathrm{a}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\mathrm{b}\right)+\left(\mathrm{x}-\mathrm{b}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\mathrm{a}\right)\end{array}$ $\begin{array}{l}\left[\mathrm{By}\text{Leibnitz rule}\right]\\ =\left(\mathrm{x}-\mathrm{a}\right)\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}-\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{b}\right)+\left(\mathrm{x}-\mathrm{b}\right)\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}-\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{a}\right)\\ =\left(\mathrm{x}-\mathrm{a}\right)\left(1-0\right)+\left(\mathrm{x}-\mathrm{b}\right)\left(1-0\right)\\ =\mathrm{x}-\mathrm{a}+\mathrm{x}-\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}-\mathrm{a}-\mathrm{b}\\ \left(\mathrm{ii}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)={\left({\mathrm{ax}}^{2}+\mathrm{b}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\right\}\\ \left[\left[\mathrm{By}\text{Leibnitz rule}\right]\right]\\ \mathrm{f}‘\left(\mathrm{x}\right)=\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\right\}\left({\mathrm{ax}}^{2}+\mathrm{b}\right)+\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{a}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{b}\right)\left({\mathrm{ax}}^{2}+\mathrm{b}\right)+\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\left(\mathrm{a}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{b}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2\mathrm{ax}+0\right)\left({\mathrm{ax}}^{2}+\mathrm{b}\right)+\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\left(2\mathrm{ax}+0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{ax}\left({\mathrm{ax}}^{2}+\mathrm{b}\right)+2\mathrm{ax}\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\mathrm{ax}\left({\mathrm{ax}}^{2}+\mathrm{b}\right)\\ \left(\mathrm{iii}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-\mathrm{a}}{\mathrm{x}-\mathrm{b}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\end{array}$ $\begin{array}{l}\text{f’}\left(\text{x}\right)=\frac{\text{d}}{\mathrm{dx}}\left(\frac{\text{x}-\text{a}}{\text{x}-\text{b}}\right)\\ \text{}=\frac{\left(\text{x}-\text{b}\right)\frac{\text{d}}{\mathrm{dx}}\left(\text{x}-\text{a}\right)-\left(\text{x}-\text{a}\right)\frac{\text{d}}{\mathrm{dx}}\left(\text{x}-\text{b}\right)}{{\left(\text{x}-\text{b}\right)}^{2}}\text{}\left[\mathrm{By}\text{quotientrule}\right]\\ \text{}=\frac{\left(\text{x}-\text{b}\right)\left(1-0\right)-\left(\text{x}-\text{a}\right)\left(1-0\right)}{{\left(\text{x}-\text{b}\right)}^{2}}\\ \text{}=\frac{\text{x}-\text{b}-\text{x}+\text{a}}{{\left(\text{x}-\text{b}\right)}^{2}}\\ \text{f’}\left(\text{x}\right)=\frac{\text{a}-\text{b}}{{\left(\text{x}-\text{b}\right)}^{2}}\end{array}$

Q.19

$\mathbf{\text{Findthederivativeof}}\frac{{\mathbf{x}}^{\mathbf{n}}\mathbf{-}{\mathbf{a}}^{\mathbf{n}}}{\mathbf{x}\mathbf{-}\mathbf{a}}\mathbf{\text{forsomeconstanta.}}$

Ans

$\begin{array}{l}\text{Wehave,}\\ \text{f}\left(\text{x}\right)=\frac{{\text{x}}^{\text{n}}-{\text{a}}^{\text{n}}}{\text{x}-\text{a}}\\ \text{Differentiatingw.r.t.x,weget}\\ \left(\text{x}\right)=\frac{\text{d}}{\mathrm{dx}}\left\{\frac{{\text{x}}^{\text{n}}-{\text{a}}^{\text{n}}}{\text{x}-\text{a}}\right\}\\ \text{}=\frac{\left(\text{x}-\text{a}\right)\frac{\text{d}}{\mathrm{dx}}\left({\text{x}}^{\text{n}}-{\text{a}}^{\text{n}}\right)-\left({\text{x}}^{\text{n}}-{\text{a}}^{\text{n}}\right)\frac{\text{d}}{\mathrm{dx}}\left(\text{x}-\text{a}\right)}{{\left(\text{x}-\text{a}\right)}^{2}}\\ \text{}\left[\mathrm{By}\text{}\mathrm{Quotient}\text{}\mathrm{Rule}\right]\\ \text{}=\frac{\left(\text{x}-\text{a}\right)\left({\text{nx}}^{\text{n}-1}-0\right)-\left({\text{x}}^{\text{n}}-{\text{a}}^{\text{n}}\right)\left(1-0\right)}{{\left(\text{x}-\text{a}\right)}^{2}}\\ \text{}=\frac{{\text{nx}}^{\text{n}}-\mathrm{an}{\text{x}}^{\text{n}-1}-{\text{x}}^{\text{n}}+{\text{a}}^{\text{n}}}{{\left(\text{x}-\text{a}\right)}^{2}}\end{array}$

Q.20

$\begin{array}{l}\mathbf{Find}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{the}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{derivative}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{of}\\ \mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{2}\mathbf{x}\mathbf{-}\frac{\mathbf{3}}{\mathbf{4}}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{\left(}\mathbf{5}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{iii}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}^{\mathbf{-}\mathbf{3}}\mathbf{\left(}\mathbf{5}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{\left(}\mathbf{iv}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{}{\mathbf{x}}^{\mathbf{5}}\mathbf{\left(}\mathbf{3}\mathbf{-}\mathbf{6}{\mathbf{x}}^{\mathbf{-}\mathbf{9}}\mathbf{\right)}\\ \mathbf{\left(}\mathbf{v}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}^{\mathbf{-}\mathbf{4}}\mathbf{\left(}\mathbf{3}\mathbf{-}\mathbf{4}{\mathbf{x}}^{\mathbf{-}\mathbf{5}}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{\left(}\mathbf{vi}\mathbf{\right)}\mathbf{}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{2}}{\mathbf{x}\mathbf{+}\mathbf{1}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{1}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{We}\mathrm{have},\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}–\frac{\mathrm{3}}{\mathrm{4}}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{2\mathrm{x}–\frac{\mathrm{3}}{\mathrm{4}}\right\}\\ =\frac{\mathrm{d}}{\mathrm{dx}}\left(2\mathrm{x}\right)\mathrm{–}\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\\ =2-0\\ =2\\ \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\left(5{\mathrm{x}}^{\mathrm{3}}+3\mathrm{x}-1\right)\left(\mathrm{x}-1\right)\end{array}$ $\begin{array}{l}\text{Differentiatingw.r.t.x,weget}\\ \text{f’}\left(\text{x}\right)\text{=}\frac{\text{d}}{\text{dx}}\left\{\left({\text{5x}}^{\text{3}}\text{+3x-1}\right)\left(\text{x-1}\right)\right\}\\ \text{=}\left\{\frac{\text{d}}{\text{dx}}\left({\text{5x}}^{\text{3}}\text{+3x-1}\right)\right\}\left(\text{x-1}\right)\text{+}\left({\text{5x}}^{\text{3}}\text{+3x-1}\right)\left\{\frac{\text{d}}{\text{dx}}\left(\text{x-1}\right)\right\}\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{}\left[\text{ByLeibnitz’sproductrule}\right]\\ \text{=}\left(\text{5}\frac{\text{d}}{\text{dx}}{\text{x}}^{\text{3}}\text{+3}\frac{\text{d}}{\text{dx}}\text{x-}\frac{\text{d}}{\text{dx}}\text{1}\right)\left(\text{x-1}\right)\text{+}\left({\text{5x}}^{\text{3}}\text{+3x-1}\right)\left(\frac{\text{d}}{\text{dx}}\text{x-}\frac{\text{d}}{\text{dx}}\text{1}\right)\\ \text{=}\left({\text{15x}}^{\text{2}}\text{+3-0}\right)\left(\text{x-1}\right)\text{+}\left({\text{5x}}^{\text{3}}\text{+3x-1}\right)\left(\text{1-0}\right)\\ {\text{=15x}}^{\text{3}}{\text{+3x-15x}}^{\text{2}}{\text{-3+5x}}^{\text{3}}\text{+3x-1}\\ {\text{=20x}}^{\text{3}}{\text{-15x}}^{\text{2}}\text{+6x-4}\\ \left(\text{iii}\right)\text{Wehave,}\\ \text{f}\left(\text{x}\right){\text{=\hspace{0.17em}x}}^{\text{-3}}\left(\text{5+3x}\right)\\ \text{Differentiatingw.r.t.x,weget}\\ \text{f’}\left(\text{x}\right)\text{=}\frac{\text{d}}{\text{dx}}\left\{{\text{x}}^{\text{-3}}\left(\text{5+3x}\right)\right\}\\ \text{=}\left\{\frac{\text{d}}{\text{dx}}{\text{x}}^{\text{-3}}\right\}\left(\text{5+3x}\right){\text{+x}}^{\text{-3}}\left\{\frac{\text{d}}{\text{dx}}\left(\text{5+3x}\right)\right\}\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{}\left[\text{By Leibnitz’s product rule}\right]\\ {\text{=-3x}}^{\text{-4}}\left(\text{5+3x}\right){\text{+x}}^{\text{-3}}\left(\text{0+3}\right)\end{array}$ $\begin{array}{l}=-15{\mathrm{x}}^{-4}-9{\mathrm{x}}^{-3}+3{\mathrm{x}}^{–3}\\ =-15{\mathrm{x}}^{-4}-6{\mathrm{x}}^{-3}\\ =-\frac{3}{{\mathrm{x}}^{4}}\left(5+2\mathrm{x}\right)\\ \left(\mathrm{iv}\right)\text{}\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}{\mathrm{x}}^{5}\left(3-6{\mathrm{x}}^{–9}\right)\\ \mathrm{Differentiating}\text{w.r.t.x,weget}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\mathrm{x}}^{5}\left(3-6{\mathrm{x}}^{–9}\right)\right\}\\ \text{}=\left\{\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{5}\right\}\left(3-6{\mathrm{x}}^{–9}\right)+{\mathrm{x}}^{5}\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(3-6{\mathrm{x}}^{–9}\right)\right\}\\ \text{}\left[\mathrm{By}\text{Leibnitz’s product rule}\right]\\ \text{}=5{\mathrm{x}}^{4}\left(3-6{\mathrm{x}}^{–9}\right)+{\mathrm{x}}^{5}\left(0+54{\mathrm{x}}^{–10}\right)\\ \text{}=15{\mathrm{x}}^{4}-30{\mathrm{x}}^{-5}+54{\mathrm{x}}^{–5}\\ \text{}=15{\mathrm{x}}^{4}+24{\mathrm{x}}^{-5}\\ \text{}=15{\mathrm{x}}^{4}+\frac{24}{{\mathrm{x}}^{5}}\\ \left(\mathrm{v}\right)\text{}\mathrm{We}\text{have,}\\ \text{}\mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}{\mathrm{x}}^{–4}\left(3-4{\mathrm{x}}^{–5}\right)\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\mathrm{x}}^{–4}\left(3-4{\mathrm{x}}^{–5}\right)\right\}\\ \text{}=\left\{\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{–4}\right\}\left(3-4{\mathrm{x}}^{–5}\right)+{\mathrm{x}}^{–4}\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(3-4{\mathrm{x}}^{–5}\right)\right\}\end{array}$ $\begin{array}{l}\left[\mathrm{By}\text{Leibnitz’sproductrule}\right]\\ \text{}=\text{}-\text{\hspace{0.17em}}4{\mathrm{x}}^{-5}\left(3-4{\mathrm{x}}^{–5}\right)+{\mathrm{x}}^{–4}\left(0+20{\mathrm{x}}^{–6}\right)\\ \text{}=\text{}-\text{\hspace{0.17em}}12{\mathrm{x}}^{-5}+16{\mathrm{x}}^{–10}+20{\mathrm{x}}^{–10}\\ \text{}=\text{}-\text{\hspace{0.17em}}12{\mathrm{x}}^{-5}+36{\mathrm{x}}^{–10}\\ \text{}=\text{}-\frac{12}{{\mathrm{x}}^{5}}+\frac{36}{{\mathrm{x}}^{10}}\\ \left(\mathrm{vi}\right)\text{}\mathrm{We}\text{have,}\\ \text{}\mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}\frac{2}{\mathrm{x}+1}-\frac{{\mathrm{x}}^{2}}{3\mathrm{x}-1}\\ \mathrm{Differentiating}\text{w.r.t.x,weget}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\text{\hspace{0.17em}}\frac{2}{\mathrm{x}+1}-\frac{{\mathrm{x}}^{2}}{3\mathrm{x}-1}\right\}\\ \text{}=\text{}\frac{\left\{\left(\mathrm{x}+1\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(2\right)-2\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+1\right)\right\}}{{\left(\mathrm{x}+1\right)}^{2}}-\frac{\left\{\left(3\mathrm{x}-1\right)\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}-{\mathrm{x}}^{2}\frac{\mathrm{d}}{\mathrm{dx}}\left(3\mathrm{x}-1\right)\right\}}{{\left(3\mathrm{x}-1\right)}^{2}}\\ \left[\mathrm{By}\text{QuotientRule}\right]\\ \text{}=\frac{\left\{\left(\mathrm{x}+1\right)\left(0\right)-2\left(1+0\right)\right\}}{{\left(\mathrm{x}+1\right)}^{2}}-\frac{\left\{\left(3\mathrm{x}-1\right)\left(2\mathrm{x}\right)-{\mathrm{x}}^{2}\left(3-0\right)\right\}}{{\left(3\mathrm{x}-1\right)}^{2}}\\ \text{}=\frac{-2}{{\left(\mathrm{x}+1\right)}^{2}}-\frac{6{\mathrm{x}}^{2}-2\mathrm{x}-3{\mathrm{x}}^{2}}{{\left(3\mathrm{x}-1\right)}^{2}}\end{array}$ $\text{=–}\frac{\text{2}}{{\left(\text{x+1}\right)}^{\text{2}}}\text{–}\frac{\text{x}\left(\text{3x-2}\right)}{{\left(\text{3x-1}\right)}^{\text{2}}}$

Q.21 Find the derivative of cos x from first principle.

Ans

$\begin{array}{l}\mathrm{Since},\text{f}\left(\mathrm{x}\right)=\mathrm{cos}\mathrm{x}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{cos}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{-2\mathrm{sin}\left(\frac{\mathrm{x}+\mathrm{h}-\mathrm{x}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{x}+\mathrm{h}+\mathrm{x}}{2}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\underset{\frac{\mathrm{h}}{2}\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\mathrm{h}}{2}\right)}{\left(\frac{\mathrm{h}}{2}\right)}×\underset{\mathrm{h}\to 0}{\mathrm{lim}}\mathrm{sin}\left(\frac{2\mathrm{x}+\mathrm{h}}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-1×\mathrm{sin}\left(\frac{2\mathrm{x}+0}{2}\right)\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{sinx}\end{array}$

Q.22 Find the derivative of the following functions:

(i) sin x cos x
(ii) sec x
(iii) 5sec x + 4cos x
(iv) cosec x
(v) 3cot x + 5cosec x
(vi) 5sin x − 6cos x + 7
(vii) 2 tan x − 7sec x

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}\mathrm{sinxcosx}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\mathrm{sinxcosx}\right\}\\ =\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}\mathrm{x}\right)\mathrm{cos}\mathrm{x}+\mathrm{sin}\mathrm{x}\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cos}\mathrm{x}\right)\\ \left[\mathrm{By}\text{Leibnitz’s product rule}\right]\\ =\mathrm{cos}\mathrm{x}\mathrm{cos}\mathrm{x}+\mathrm{sin}\mathrm{x}\left(-\mathrm{sin}\mathrm{x}\right)\\ ={\mathrm{cos}}^{2}\mathrm{x}-{\mathrm{sin}}^{2}\mathrm{x}=\mathrm{cos}2\mathrm{x}\\ \left(\mathrm{ii}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}\mathrm{secx}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sec}\mathrm{x}\right)\\ =\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}\\ \left(\mathrm{iii}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}5\mathrm{sec}\mathrm{x}+4\mathrm{cos}\mathrm{x}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(5\mathrm{sec}\mathrm{x}+4\mathrm{cos}\mathrm{x}\right)\\ =5\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sec}\mathrm{x}+4\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cos}\mathrm{x}\\ =5\left(\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}\right)+4\left(-\mathrm{sin}\mathrm{x}\right)\\ =5\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}-4\mathrm{sin}\mathrm{x}\end{array}$ $\begin{array}{l}\left(\mathrm{iv}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}\mathrm{cos}\text{e}\mathrm{c}\mathrm{x}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\text{cosec}\mathrm{x}\right)\\ =-\text{cosec}\mathrm{x}\mathrm{cot}\mathrm{x}\\ \left(\mathrm{v}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}3\mathrm{cot}\mathrm{x}+5\mathrm{cosec}\mathrm{x}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(3\mathrm{cot}\mathrm{x}+5\mathrm{cosec}\mathrm{x}\right)\\ =3\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cot}\mathrm{x}+5\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosec}\mathrm{x}\\ =3\left(-{\mathrm{cosec}}^{2}\mathrm{x}\right)-5\mathrm{cosec}\mathrm{x}\mathrm{cot}\mathrm{x}\\ =-3{\mathrm{cosec}}^{2}\mathrm{x}-5\mathrm{cosec}\mathrm{x}\mathrm{cot}\mathrm{x}\\ \left(\mathrm{vi}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=\text{\hspace{0.17em}}5\mathrm{sin}\mathrm{x}-6\mathrm{cos}\mathrm{x}+7\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(5\mathrm{sin}\mathrm{x}-6\mathrm{cos}\mathrm{x}+7\right)\\ =5\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}\mathrm{x}-6\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cos}\mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}}7\\ =5\mathrm{cos}\mathrm{x}+6\mathrm{sin}\mathrm{x}+0\\ =5\mathrm{cos}\mathrm{x}+6\mathrm{sin}\mathrm{x}\\ \left(\mathrm{vii}\right)\mathrm{We}\text{have,}\\ \mathrm{f}\left(\mathrm{x}\right)=2\mathrm{tan}\mathrm{x}-7\mathrm{sec}\mathrm{x}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{Differentiating}\text{w.r.t. x, we get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(2\mathrm{tan}\mathrm{x}-7\mathrm{sec}\mathrm{x}\right)\\ =2\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{tan}\mathrm{x}-7\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sec}\mathrm{x}\\ =2{\mathrm{sec}}^{2}\mathrm{x}-7\mathrm{secx}\mathrm{tan}\mathrm{x}\end{array}$

Q.23

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{4}\mathbf{x}}\mathbf{-}\mathbf{1}}{\mathbf{x}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{4\mathrm{x}}-1}{\mathrm{x}}=\underset{4\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{4\mathrm{x}}-1}{4\mathrm{x}}×4\left(\mathrm{x}\to 0⇒4\mathrm{x}\to 0\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=1×4\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=4\end{array}$

Q.24

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{2}\mathbf{+}\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{2}}}{\mathbf{x}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{2+\mathrm{x}}-{\mathrm{e}}^{2}}{\mathrm{x}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{2}\left({\mathrm{e}}^{\mathrm{x}}-1\right)}{\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{2}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\left({\mathrm{e}}^{\mathrm{x}}-1\right)}{\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{2}×1\left[âˆµ\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{2}\end{array}$

Q.25

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{5}}}{\mathbf{x}\mathbf{-}\mathbf{5}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{5}}{\mathrm{x}-5}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{5}\left({\mathrm{e}}^{\mathrm{x}-5}-1\right)}{\mathrm{x}-5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{5}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{x}-5}-1}{\mathrm{x}-5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{5}×1\left[âˆµ\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{5}\end{array}$

Q.26

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{sinx}}\mathbf{-}\mathbf{1}}{\mathbf{x}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{sinx}}-1}{\mathrm{x}}=\underset{\mathrm{sinx}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{sinx}}-1}{\mathrm{sinx}}×\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{sinx}}{\mathrm{x}}\left(\mathrm{x}\to 0⇒\mathrm{sinx}\to 0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1×1\left[\begin{array}{l}âˆµ\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\text{and}\\ \text{}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{sinx}}{\mathrm{x}}=1\end{array}\right]\\ \therefore \underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{sinx}}-1}{\mathrm{x}}=1\end{array}$

Q.27

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{3}}}{\mathbf{x}\mathbf{-}\mathbf{3}}$

Ans

$\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{3}}{\mathrm{x}-3}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{3}\left({\mathrm{e}}^{\mathrm{x}-3}-1\right)}{\mathrm{x}-3}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{3}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{x}-3}-1}{\mathrm{x}-3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{3}×1\left[âˆµ\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{e}}^{3}\end{array}$

Q.28

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{x}\mathbf{\left(}{\mathbf{e}}^{\mathbf{x}}\mathbf{-}\mathbf{1}\mathbf{\right)}}{\mathbf{1}\mathbf{-}\mathbf{cosx}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{x}\left({\mathrm{e}}^{\mathrm{x}}-1\right)}{1-\mathrm{cosx}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}×\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{x}}^{2}}{1-\mathrm{cosx}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1×\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{x}}^{2}}{2{\mathrm{sin}}^{2}\left(\frac{\mathrm{x}}{2}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{2{\left(\frac{\mathrm{x}}{2}\right)}^{2}}{{\left\{\mathrm{sin}\left(\frac{\mathrm{x}}{2}\right)\right\}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{2}{{\left\{\frac{\mathrm{sin}\left(\frac{\mathrm{x}}{2}\right)}{\left(\frac{\mathrm{x}}{2}\right)}\right\}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{{\left(1\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\end{array}$

Q.29

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{log}}_{\mathbf{e}}\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{log}}_{\mathrm{e}}\left(1+2\mathrm{x}\right)}{\mathrm{x}}=\underset{2\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{log}}_{\mathrm{e}}\left(1+2\mathrm{x}\right)}{2\mathrm{x}}×2\left[\mathrm{x}\to 0⇒2\mathrm{x}\to 0\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1×2\left[âˆµ\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{{\mathrm{log}}_{\mathrm{e}}\left(1+\mathrm{x}\right)}{\mathrm{x}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\end{array}$

Q.30

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{log}\mathbf{\left(}\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{3}}\mathbf{\right)}}{{\mathbf{sin}}^{\mathbf{3}}\mathbf{x}}$

Ans

$\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1+{\mathrm{x}}^{3}\right)}{{\mathrm{sin}}^{3}\mathrm{x}}=\underset{{\mathrm{x}}^{3}\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1+{\mathrm{x}}^{3}\right)}{{\mathrm{x}}^{3}}×\underset{{\mathrm{x}}^{3}\to 0}{\mathrm{lim}}\frac{{\mathrm{x}}^{3}}{{\mathrm{sin}}^{3}\mathrm{x}}\text{\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\mathrm{x}\to 0⇒{\mathrm{x}}^{3}\to 0\right]$ $\begin{array}{l}=1×1\left[âˆµ\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1+\mathrm{x}\right)}{\mathrm{x}}=1\text{and}\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{x}}{\mathrm{sinx}}=1\right]\\ =1\end{array}$

Q.31 Find the derivative of the following functions from first principle:

$\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{}\mathbf{-}\mathbf{x}\mathbf{}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{}{\mathbf{\left(}\mathbf{-}\mathbf{x}\mathbf{\right)}}^{\mathbf{–}\mathbf{1}}\mathbf{}\mathbf{\left(}\mathbf{iii}\mathbf{\right)}\mathbf{}\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{}\mathbf{\left(}\mathbf{iv}\mathbf{\right)}\mathbf{}\mathbf{cos}\mathbf{\left(}\mathbf{x}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{8}}\mathbf{\right)}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Since},\text{f}\left(\mathrm{x}\right)=-\text{\hspace{0.17em}}\mathrm{x}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{-\left(\mathrm{x}+\mathrm{h}\right)-\left(-\mathrm{x}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{-\mathrm{x}-\mathrm{h}+\mathrm{x}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{h}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-1\\ \left(\mathrm{ii}\right)\mathrm{Since},\text{f}\left(\mathrm{x}\right)={\left(-\text{\hspace{0.17em}}\mathrm{x}\right)}^{-1}=\frac{1}{-\mathrm{x}}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{-\frac{1}{\left(\mathrm{x}+\mathrm{h}\right)}-\left(\frac{1}{-\mathrm{x}}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{-\frac{1}{\left(\mathrm{x}+\mathrm{h}\right)}+\frac{1}{\mathrm{x}}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\frac{-\mathrm{x}+\mathrm{x}+\mathrm{h}}{\mathrm{x}\left(\mathrm{x}+\mathrm{h}\right)}}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{h}}{\mathrm{hx}\left(\mathrm{x}+\mathrm{h}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{x}\left(\mathrm{x}+0\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{-2}\\ \left(\mathrm{iii}\right)\mathrm{Since},\text{f}\left(\mathrm{x}\right)=\mathrm{sin}\left(\mathrm{x}+1\right)\end{array}$ $\begin{array}{l}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\mathrm{x}+\mathrm{h}+1\right)-\mathrm{sin}\left(\mathrm{x}+1\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{2\mathrm{cos}\left(\frac{\mathrm{x}+\mathrm{h}+1+\mathrm{x}+1}{2}\right)\mathrm{sin}\left(\frac{\mathrm{x}+\mathrm{h}+1-\mathrm{x}-1}{2}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{2\mathrm{cos}\left(\frac{2\mathrm{x}+\mathrm{h}+2}{2}\right)\mathrm{sin}\left(\frac{\mathrm{h}}{2}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\mathrm{cos}\left(\frac{2\mathrm{x}+\mathrm{h}+2}{2}\right)×\underset{\frac{\mathrm{h}}{2}\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\mathrm{h}}{2}\right)}{\left(\frac{\mathrm{h}}{2}\right)}\left[\mathrm{h}\to 0⇒\frac{\mathrm{h}}{2}\to 0\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cos}\left(\frac{2\mathrm{x}+0+2}{2}\right)×1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cos}\left(\mathrm{x}+1\right)\\ \left(\mathrm{iv}\right)\mathrm{Since},\text{f}\left(\mathrm{x}\right)=\mathrm{cos}\left(\mathrm{x}-\frac{\mathrm{\pi }}{8}\right)\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{cos}\left(\mathrm{x}+\mathrm{h}-\frac{\mathrm{\pi }}{8}\right)-\mathrm{cos}\left(\mathrm{x}-\frac{\mathrm{\pi }}{8}\right)}{\mathrm{h}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{-2\mathrm{sin}\left(\frac{\mathrm{x}+\mathrm{h}-\frac{\mathrm{\pi }}{8}+\mathrm{x}-\frac{\mathrm{\pi }}{8}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{x}+\mathrm{h}-\frac{\mathrm{\pi }}{8}-\mathrm{x}+\frac{\mathrm{\pi }}{8}}{2}\right)}{\mathrm{h}}\end{array}$ $\begin{array}{l}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=-\underset{\mathrm{h}\to 0}{\mathrm{lim}}\mathrm{sin}\left(\frac{2\mathrm{x}+\mathrm{h}-\frac{2\mathrm{\pi }}{8}}{2}\right)×\underset{\frac{\mathrm{h}}{2}\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\mathrm{h}}{2}\right)}{\left(\frac{\mathrm{h}}{2}\right)}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{h}\to 0⇒\frac{\mathrm{h}}{2}\to 0\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{sin}\left(\frac{2\mathrm{x}+0-\frac{2\mathrm{\pi }}{8}}{2}\right)×1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{sin}\left(\mathrm{x}-\frac{\mathrm{\pi }}{8}\right)\end{array}$

Q.32 Find the derivative of the function (x + a).

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{a}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\mathrm{a}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1+0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1\end{array}$

Q.33

$\mathbf{\text{Find the derivative of the function}}\left(\text{px+q}\right)\left(\frac{\text{r}}{\text{x}}\text{+s}\right)\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\left(\mathrm{px}+\mathrm{q}\right)\left(\frac{\mathrm{r}}{\mathrm{x}}+\mathrm{s}\right)\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\end{array}$ $\begin{array}{l}\text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left(\mathrm{px}+\mathrm{q}\right)\left(\frac{\mathrm{r}}{\mathrm{x}}+\mathrm{s}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{px}+\mathrm{q}\right)\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{r}}{\mathrm{x}}+\mathrm{s}\right)\right\}+\left(\frac{\mathrm{r}}{\mathrm{x}}+\mathrm{s}\right)\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{px}+\mathrm{q}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{px}+\mathrm{q}\right)\left(\mathrm{r}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{-1}+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{s}\right)+\left(\frac{\mathrm{r}}{\mathrm{x}}+\mathrm{s}\right)\left(\mathrm{p}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{q}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{px}+\mathrm{q}\right)\left(-{\mathrm{rx}}^{-2}+0\right)+\left(\frac{\mathrm{r}}{\mathrm{x}}+\mathrm{s}\right)\left(\mathrm{p}×1+0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{\mathrm{pr}}{\mathrm{x}}-{\mathrm{qrx}}^{-2}+\frac{\mathrm{pr}}{\mathrm{x}}+\mathrm{ps}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-{\mathrm{qrx}}^{-2}+\mathrm{ps}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{\mathrm{qr}}{{\mathrm{x}}^{2}}+\mathrm{ps}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=-\frac{\mathrm{qr}}{{\mathrm{x}}^{2}}+\mathrm{ps}\end{array}$

Q.34 Find the derivative of the function (ax + b)(cx + d)2.

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\left(\mathrm{ax}\text{}+\text{}\mathrm{b}\right){\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)}^{2}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left(\mathrm{ax}\text{}+\text{}\mathrm{b}\right){\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)}^{2}\right\}\left[\mathrm{By}\text{Leibnitz product rule}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{ax}\text{}+\text{}\mathrm{b}\right)\left\{\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)}^{2}\right\}+{\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)}^{2}\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}\text{}+\text{}\mathrm{b}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{ax}\text{}+\text{}\mathrm{b}\right)\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)\right\}+{\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)}^{2}\left(\mathrm{a}+0\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{ax}\text{}+\text{}\mathrm{b}\right)\left\{\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)+\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)\right\}\\ +\mathrm{a}{\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{ax}\text{}+\text{}\mathrm{b}\right)\left\{\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)\left(\mathrm{c}+0\right)+\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)\left(\mathrm{c}+0\right)\right\}\\ +\mathrm{a}{\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{ax}\text{}+\text{}\mathrm{b}\right)2\mathrm{c}\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)+\mathrm{a}{\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{c}\left(\mathrm{ax}\text{}+\text{}\mathrm{b}\right)\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)+\mathrm{a}{\left(\mathrm{cx}\text{}+\text{}\mathrm{d}\right)}^{2}\end{array}$

Q.35

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{ax+b}}}{\mathbf{\text{cx+d}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+\mathrm{d}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+\mathrm{d}}\right\}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{cx}+\mathrm{d}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}+\mathrm{b}\right)-\left(\mathrm{ax}+\mathrm{b}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{cx}+\mathrm{d}\right)}{{\left(\mathrm{cx}+\mathrm{d}\right)}^{2}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{cx}+\mathrm{d}\right)\left(\mathrm{a}+0\right)-\left(\mathrm{ax}+\mathrm{b}\right)\left(\mathrm{c}+0\right)}{{\left(\mathrm{cx}+\mathrm{d}\right)}^{2}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{a}\left(\mathrm{cx}+\mathrm{d}\right)-\mathrm{c}\left(\mathrm{ax}+\mathrm{b}\right)}{{\left(\mathrm{cx}+\mathrm{d}\right)}^{2}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{acx}+\mathrm{ad}-\mathrm{acx}-\mathrm{bc}}{{\left(\mathrm{cx}+\mathrm{d}\right)}^{2}}=\frac{\mathrm{ad}-\mathrm{bc}}{{\left(\mathrm{cx}+\mathrm{d}\right)}^{2}}\end{array}$

Q.36

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{1+}}\frac{\mathbf{\text{1}}}{\mathbf{\text{x}}}}{\mathbf{\text{1}}\mathbf{-}\frac{\mathbf{\text{1}}}{\mathbf{\text{x}}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{1+\frac{1}{\mathrm{x}}}{1-\frac{1}{\mathrm{x}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}+1}{\mathrm{x}-1}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}-1\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+1\right)-\left(\mathrm{x}+1\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-1\right)}{{\left(\mathrm{x}-1\right)}^{2}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}-1\right)\left(1+0\right)-\left(\mathrm{x}+1\right)\left(1-0\right)}{{\left(\mathrm{x}-1\right)}^{2}}\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}-1-\mathrm{x}-1}{{\left(\mathrm{x}-1\right)}^{2}}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{-2}{{\left(\mathrm{x}-1\right)}^{2}}\end{array}$

Q.37

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{1}}}{{\mathbf{\text{ax}}}^{\mathbf{\text{2}}}\mathbf{\text{+ bx + c}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\end{array}$ $\begin{array}{l}\text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{{\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left({\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(1\right)-\left(1\right)\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}\right)}{{\left({\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left({\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}\right)\left(0\right)-\left(1\right)\left(2\mathrm{ax}+\mathrm{b}+0\right)}{{\left({\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-\left(2\mathrm{ax}+\mathrm{b}\right)}{{\left({\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}\right)}^{2}}\end{array}$

Q.38

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{ax + b}}}{{\mathbf{\text{px}}}^{\mathbf{\text{2}}}\mathbf{\text{+ qx + r}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}+\mathrm{b}\right)-\left(\mathrm{ax}+\mathrm{b}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)}{{\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)\left(\mathrm{a}\right)-\left(\mathrm{ax}+\mathrm{b}\right)\left(2\mathrm{px}+\mathrm{q}+0\right)}{{\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)}^{2}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{a}\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)-\left(\mathrm{ax}+\mathrm{b}\right)\left(2\mathrm{px}+\mathrm{q}\right)}{{\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)}^{2}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{apx}}^{2}+\mathrm{aqx}+\mathrm{ar}-\left(2{\mathrm{apx}}^{2}+\mathrm{axq}+2\mathrm{bpx}+\mathrm{bq}\right)}{{\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)}^{2}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{apx}}^{2}+\mathrm{aqx}+\mathrm{ar}-2{\mathrm{apx}}^{2}-\mathrm{axq}-2\mathrm{bpx}-\mathrm{bq}}{{\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)}^{2}}\\ \text{f’}\left(\mathrm{x}\right)=\frac{-{\mathrm{apx}}^{2}-2\mathrm{bpx}+\mathrm{ar}-\mathrm{bq}}{{\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)}^{2}}\end{array}$

Q.39

$\mathbf{\text{Find the derivative of the function}}\frac{{\mathbf{\text{px}}}^{\mathbf{\text{2}}}\mathbf{\text{+ qx + r}}}{\mathbf{\text{ax + b}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{{\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}}{\mathrm{ax}+\mathrm{b}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}}{\mathrm{ax}+\mathrm{b}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{ax}+\mathrm{b}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)-\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}+\mathrm{b}\right)}{{\left(\mathrm{ax}+\mathrm{b}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{ax}+\mathrm{b}\right)\left(2\mathrm{px}+\mathrm{q}+0\right)-\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)\left(\mathrm{a}\right)}{{\left(\mathrm{ax}+\mathrm{b}\right)}^{2}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{ax}+\mathrm{b}\right)\left(2\mathrm{px}+\mathrm{q}\right)-\mathrm{a}\left({\mathrm{px}}^{2}+\mathrm{qx}+\mathrm{r}\right)}{{\left(\mathrm{ax}+\mathrm{b}\right)}^{2}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2{\mathrm{apx}}^{2}+\mathrm{axq}+2\mathrm{bpx}+\mathrm{bq}-\left({\mathrm{apx}}^{2}+\mathrm{aqx}+\mathrm{ar}\right)}{{\left(\mathrm{ax}+\mathrm{b}\right)}^{2}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2{\mathrm{apx}}^{2}+\mathrm{axq}+2\mathrm{bpx}+\mathrm{bq}-{\mathrm{apx}}^{2}-\mathrm{aqx}-\mathrm{ar}}{{\left(\mathrm{ax}+\mathrm{b}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}f’}\left(\mathrm{x}\right)=\frac{{\mathrm{apx}}^{2}+2\mathrm{bpx}+\mathrm{bq}-\mathrm{ar}}{{\left(\mathrm{ax}+\mathrm{b}\right)}^{2}}\end{array}$

Q.40

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{a}}}{{\mathbf{\text{x}}}^{\mathbf{\text{4}}}}\mathbf{-}\frac{\mathbf{\text{b}}}{{\mathbf{\text{x}}}^{\mathbf{\text{2}}}}\mathbf{\text{+cos x}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{{\mathrm{x}}^{4}}-\frac{\mathrm{b}}{{\mathrm{x}}^{2}}+\mathrm{cosx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{ax}}^{-4}-{\mathrm{bx}}^{-2}+\mathrm{cosx}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{ax}}^{-4}-{\mathrm{bx}}^{-2}+\mathrm{cosx}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{a}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{-4}-\mathrm{b}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{-2}+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cos}\mathrm{x}\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}4{\mathrm{ax}}^{-5}+2{\mathrm{bx}}^{-3}-\mathrm{sin}\mathrm{x}\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}\frac{4\mathrm{a}}{{\mathrm{x}}^{5}}+\frac{2\mathrm{b}}{{\mathrm{x}}^{3}}-\mathrm{sin}\mathrm{x}\end{array}$

Q.41

$\mathbf{\text{Find the derivative of the function 4}}\sqrt{\mathbf{\text{x}}}\mathbf{-}\mathbf{\text{2.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=4\sqrt{\mathrm{x}}-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4{\mathrm{x}}^{\frac{1}{2}}-2\end{array}$ $\begin{array}{l}\mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{\hspace{0.17em}f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(4{\mathrm{x}}^{\frac{1}{2}}-2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=4\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{\frac{1}{2}}-\frac{\mathrm{d}}{\mathrm{dx}}\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4×\frac{1}{2}{\mathrm{x}}^{-\frac{1}{2}}-0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\text{\hspace{0.17em}}{\mathrm{x}}^{-\frac{1}{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{\sqrt{\mathrm{x}}}\end{array}$

Q.42 Find the derivative of the function (ax + b)n.

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)={\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\text{and f}\left(\mathrm{x}+\mathrm{h}\right)={\left\{\mathrm{a}\left(\mathrm{x}+\mathrm{h}\right)+\mathrm{b}\right\}}^{\mathrm{n}}\\ \text{}={\left(\mathrm{ax}+\mathrm{ah}+\mathrm{b}\right)}^{\mathrm{n}}\\ \mathrm{By}\text{first Principle,\hspace{0.17em}\hspace{0.17em}f’}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \text{}\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{ax}+\mathrm{ah}+\mathrm{b}\right)}^{\mathrm{n}}-{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}}{\mathrm{h}}\\ \text{}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\text{}\frac{{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{{\left(\frac{\mathrm{ax}+\mathrm{ah}+\mathrm{b}}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{n}}-1\right\}}{\mathrm{h}}\\ \text{}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{{\left(\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{ax}+\mathrm{b}}+\frac{\mathrm{ah}}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{n}}-1\right\}}{\mathrm{h}}\\ \text{}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{{\left(1+\frac{\mathrm{ah}}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{n}}-1\right\}}{\mathrm{h}}\\ \text{}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{1+\frac{\mathrm{nah}}{\mathrm{ax}+\mathrm{b}}+\frac{\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}^{2}{\mathrm{h}}^{2}}{2!\left(\mathrm{ax}+\mathrm{b}\right)}+...-1\right\}}{\mathrm{h}}\\ \text{}\left[\mathrm{Using}\text{binomial theorem}\right]\\ \text{}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{\frac{\mathrm{na}}{\mathrm{ax}+\mathrm{b}}+\frac{\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}^{2}\mathrm{h}}{2!\left(\mathrm{ax}+\mathrm{b}\right)}+...\right\}\\ \text{}={\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{\frac{\mathrm{na}}{\mathrm{ax}+\mathrm{b}}+0+...\right\}\\ \text{}=\frac{\mathrm{na}}{\mathrm{ax}+\mathrm{b}}×{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\\ \text{\hspace{0.17em}}\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\text{\hspace{0.17em}}=\mathrm{na}{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}-1}\end{array}$

Q.43 Find the derivative of the function (ax + b)n (cx + d)m.

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)={\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\right\}\\ \mathrm{By}\text{Leibnitz product rule:}\\ \text{f’}\left(\mathrm{x}\right)={\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}+{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\\ ={\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}_{2}+{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}_{1}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\end{array}$ $\begin{array}{l}\left[\mathrm{Let}{\text{y}}_{\text{1}}={\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}},{\text{y}}_{\text{2}}={\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\right]\\ \mathrm{By}\text{first principle:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}_{1}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{ax}+\mathrm{ah}+\mathrm{b}\right)}^{\mathrm{n}}-{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{{\left(\frac{\mathrm{ax}+\mathrm{ah}+\mathrm{b}}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{n}}-1\right\}}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{{\left(\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{ax}+\mathrm{b}}+\frac{\mathrm{ah}}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{n}}-1\right\}}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{{\left(1+\frac{\mathrm{ah}}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{n}}-1\right\}}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{1+\frac{\mathrm{nah}}{\mathrm{ax}+\mathrm{b}}+\frac{\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}^{2}{\mathrm{h}}^{2}}{2!\left(\mathrm{ax}+\mathrm{b}\right)}+...-1\right\}}{\mathrm{h}}\\ \left[\mathrm{Using}\text{binomial theorem}\right]\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{\frac{\mathrm{na}}{\mathrm{ax}+\mathrm{b}}+\frac{\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}^{2}\mathrm{h}}{2!\left(\mathrm{ax}+\mathrm{b}\right)}+...\right\}\\ ={\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\left\{\frac{\mathrm{na}}{\mathrm{ax}+\mathrm{b}}+0+...\right\}\\ =\frac{\mathrm{na}}{\mathrm{ax}+\mathrm{b}}×{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}}\\ \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}_{1}=\mathrm{na}{\left(\mathrm{ax}+\mathrm{b}\right)}^{\mathrm{n}-1}\end{array}$ $\begin{array}{l}\mathrm{and}\text{\hspace{0.17em}}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}_{2}=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{cx}+\mathrm{ch}+\mathrm{d}\right)}^{\mathrm{m}}-{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\left\{{\left(\frac{\mathrm{cx}+\mathrm{ch}+\mathrm{d}}{\mathrm{cx}+\mathrm{d}}\right)}^{\mathrm{m}}-1\right\}}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\left\{{\left(\frac{\mathrm{cx}+\mathrm{d}}{\mathrm{cx}+\mathrm{d}}+\frac{\mathrm{ch}}{\mathrm{cx}+\mathrm{d}}\right)}^{\mathrm{m}}-1\right\}}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\left\{{\left(1+\frac{\mathrm{ch}}{\mathrm{cx}+\mathrm{d}}\right)}^{\mathrm{m}}-1\right\}}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\left\{1+\frac{\mathrm{mch}}{\mathrm{cx}+\mathrm{d}}+\frac{\mathrm{m}\left(\mathrm{m}-1\right){\mathrm{c}}^{2}{\mathrm{h}}^{2}}{2!\left(\mathrm{cx}+\mathrm{d}\right)}+...-1\right\}}{\mathrm{h}}\\ \left[\mathrm{Using}\text{binomial theorem}\right]\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\left\{\frac{\mathrm{mc}}{\mathrm{cx}+\mathrm{d}}+\frac{\mathrm{m}\left(\mathrm{m}-1\right){\mathrm{c}}^{2}\mathrm{h}}{2!\left(\mathrm{cx}+\mathrm{d}\right)}+...\right\}\\ ={\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\left\{\frac{\mathrm{mc}}{\mathrm{cx}+\mathrm{d}}+0+...\right\}\\ =\frac{\mathrm{mc}}{\mathrm{cx}+\mathrm{d}}×{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}}\\ \frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}_{2}=\mathrm{mc}{\left(\mathrm{cx}+\mathrm{d}\right)}^{\mathrm{m}-1}\\ \mathrm{Substituting}\text{the values of}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}_{1}\text{\hspace{0.17em}}\mathrm{and}\text{}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}_{2}\text{}\mathrm{in}\text{}\mathrm{equation}\left(\mathrm{i}\right),\\ \mathrm{we}\text{}\mathrm{get}\end{array}$ $\begin{array}{l}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\end{array}$ $\begin{array}{l}\left(\text{x}\right)={\left(\mathrm{ax}+\text{b}\right)}^{\text{n}}\mathrm{mc}{\left(\mathrm{cx}+\text{d}\right)}^{\text{m}-1}+{\left(\mathrm{cx}+\text{d}\right)}^{\text{m}}\mathrm{na}{\left(\mathrm{ax}+\text{b}\right)}^{\text{n}-1}\\ \text{}={\left(\mathrm{ax}+\text{b}\right)}^{\text{n}-1}{\left(\mathrm{cx}+\text{d}\right)}^{\text{m}-1}\left\{\mathrm{mc}\left(\mathrm{ax}+\text{b}\right)+\mathrm{na}\left(\mathrm{cx}+\text{d}\right)\right\}\end{array}$

Q.44 Find the derivative of the function sin(x + a).

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)\text{\hspace{0.17em}and f}\left(\mathrm{x}+\mathrm{h}\right)=\mathrm{sin}\left(\mathrm{x}+\mathrm{h}+\mathrm{a}\right)\\ \mathrm{By}\text{first Principle,}\\ \text{f’}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\mathrm{x}+\mathrm{h}+\mathrm{a}\right)-\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\frac{2\mathrm{cos}\left(\frac{\mathrm{x}+\mathrm{h}+\mathrm{a}+\mathrm{x}+\mathrm{a}}{2}\right)\mathrm{sin}\left(\frac{\mathrm{x}+\mathrm{h}+\mathrm{a}-\mathrm{x}-\mathrm{a}}{2}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\mathrm{lim}}\mathrm{cos}\left(\frac{2\mathrm{x}+\mathrm{h}+2\mathrm{a}}{2}\right)×\underset{\frac{\mathrm{h}}{2}\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\mathrm{h}}{2}\right)}{\left(\frac{\mathrm{h}}{2}\right)}\left[\mathrm{h}\to 0⇒\frac{\mathrm{h}}{2}\to 0\right]\end{array}$ $\begin{array}{l}=\mathrm{cos}\left(\frac{2\mathrm{x}+0+2\mathrm{a}}{2}\right)×1\\ =\mathrm{cos}\left(\mathrm{x}+\mathrm{a}\right)×1\\ =\mathrm{cos}\left(\mathrm{x}+\mathrm{a}\right)\\ \mathrm{Therefore},\\ \frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)=\mathrm{cos}\left(\mathrm{x}+\mathrm{a}\right)\end{array}$

Q.45 Find the derivative of the function cosec x cot x.

Ans

$\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}\text{}\mathrm{cot}\mathrm{x}$ $\begin{array}{l}\mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}\text{}\mathrm{cot}\mathrm{x}\right)\\ \mathrm{By}\text{Leibnitz product rule:}\\ \text{f’}\left(\mathrm{x}\right)=\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cot}\mathrm{x}+\mathrm{cot}\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}\\ =\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}\left(-{\mathrm{cosec}}^{2}\mathrm{x}\right)+\mathrm{cot}\mathrm{x}\left(-\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}\mathrm{cot}\mathrm{x}\right)\\ =-{\mathrm{cosec}}^{3}\text{\hspace{0.17em}}\mathrm{x}-\mathrm{cosec}\text{\hspace{0.17em}}\mathrm{x}{\mathrm{cot}}^{2}\mathrm{x}\end{array}$

Q.46

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{cos x}}}{\mathbf{\text{1+sin x}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{\mathrm{cosx}}{1+\mathrm{sinx}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{cosx}}{1+\mathrm{sinx}}\right)\\ =\frac{\left(1+\mathrm{sin}\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cos}\mathrm{x}-\mathrm{cos}\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+\mathrm{sin}\mathrm{x}\right)}{{\left(1+\mathrm{sin}\mathrm{x}\right)}^{2}}\\ =\frac{\left(1+\mathrm{sin}\mathrm{x}\right)\left(-\mathrm{sin}\mathrm{x}\right)-\mathrm{cos}\mathrm{x}\left(0+\mathrm{cos}\mathrm{x}\right)}{{\left(1+\mathrm{sin}\mathrm{x}\right)}^{2}}\\ =\frac{-\mathrm{sin}\mathrm{x}-{\mathrm{sin}}^{2}\mathrm{x}-{\mathrm{cos}}^{2}\mathrm{x}}{{\left(1+\mathrm{sin}\mathrm{x}\right)}^{2}}\\ =\frac{-\mathrm{sin}\mathrm{x}-\left({\mathrm{sin}}^{2}\mathrm{x}+{\mathrm{cos}}^{2}\mathrm{x}\right)}{{\left(1+\mathrm{sinx}\right)}^{2}}\end{array}$ $\begin{array}{l}=\frac{-\mathrm{sin}\mathrm{x}-1}{{\left(1+\mathrm{sin}\mathrm{x}\right)}^{2}}\\ =\frac{-\left(1+\mathrm{sin}\mathrm{x}\right)}{{\left(1+\mathrm{sin}\mathrm{x}\right)}^{2}}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{-1}{\left(1+\mathrm{sin}\mathrm{x}\right)}\end{array}$

Q.47

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{sinx + cosx}}}{\mathbf{\text{sinx}}\mathbf{-}\mathbf{\text{cosx}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{\mathrm{sin}\mathrm{x}+\mathrm{cos}\mathrm{x}}{\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{sin}\mathrm{x}+\mathrm{cos}\mathrm{x}}{\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}}\right)\\ =\frac{\left\{\begin{array}{l}\left(\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sin}\mathrm{x}+\mathrm{cos}\mathrm{x}\right)\\ -\left(\mathrm{sin}\mathrm{x}+\mathrm{cos}\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}\right)\end{array}\right\}}{{\left(\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}\right)}^{2}}\\ =\frac{\left(\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}\right)\left(\mathrm{cos}\mathrm{x}-\mathrm{sin}\mathrm{x}\right)-\left(\mathrm{sin}\mathrm{x}+\mathrm{cos}\mathrm{x}\right)\left(\mathrm{cos}\mathrm{x}+\mathrm{sin}\mathrm{x}\right)}{{\left(\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}\right)}^{2}}\\ =\frac{-\left({\mathrm{cos}}^{2}\mathrm{x}-2\mathrm{cos}\mathrm{x}\mathrm{sin}\mathrm{x}+{\mathrm{sin}}^{2}\mathrm{x}\right)-\left(\begin{array}{l}{\mathrm{sin}}^{2}\mathrm{x}+2\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}\\ +{\mathrm{cos}}^{2}\mathrm{x}\end{array}\right)}{{\left(\mathrm{sinx}-\mathrm{cosx}\right)}^{2}}\end{array}$ $\begin{array}{l}=\frac{-\left(1-2\mathrm{cos}\mathrm{x}\mathrm{sin}\mathrm{x}\right)-\left(1+2\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}\right)}{{\left(\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}\right)}^{2}}\\ =\frac{-1+2\mathrm{cos}\mathrm{x}\mathrm{sin}\mathrm{x}-1-2\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}}{{\left(\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}f’}\left(\mathrm{x}\right)=\frac{-2}{{\left(\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}\right)}^{2}}\end{array}$

Q.48

$\text{Find the derivative of the function}\frac{\text{secx}-\text{1}}{\text{secx+1}}\text{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{\mathrm{sec}\mathrm{x}-1}{\mathrm{sec}\mathrm{x}+1}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{sec}\mathrm{x}-1}{\mathrm{sec}\mathrm{x}+1}\right)\end{array}$ $\begin{array}{l}=\frac{\left\{\left(\mathrm{sec}\mathrm{x}+1\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sec}\mathrm{x}-1\right)-\left(\mathrm{sec}\mathrm{x}-1\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sec}\mathrm{x}+1\right)\right\}}{{\left(\mathrm{sec}\mathrm{x}+1\right)}^{2}}\\ =\frac{\left(\mathrm{sec}\mathrm{x}+1\right)\left(\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}-0\right)-\left(\mathrm{sec}\mathrm{x}-1\right)\left(\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}+0\right)}{{\left(\mathrm{sec}\mathrm{x}+1\right)}^{2}}\\ =\frac{\left({\mathrm{sec}}^{2}\mathrm{x}\mathrm{tan}\mathrm{x}+\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}\right)-\left({\mathrm{sec}}^{2}\mathrm{x}\mathrm{tan}\mathrm{x}-\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}\right)}{{\left(\mathrm{sec}\mathrm{x}+1\right)}^{2}}\\ =\frac{{\mathrm{sec}}^{2}\mathrm{x}\mathrm{tan}\mathrm{x}+\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}-{\mathrm{sec}}^{2}\mathrm{x}\mathrm{tan}\mathrm{x}+\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}}{{\left(\mathrm{sec}\mathrm{x}+1\right)}^{2}}\\ \text{f’}\left(\mathrm{x}\right)=\frac{2\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}}{{\left(\mathrm{sec}\mathrm{x}+1\right)}^{2}}\end{array}$

Q.49 Find the derivative of the function sinn x.

Ans

$\begin{array}{l}\mathrm{Let}\text{y}={\mathrm{sin}}^{\mathrm{n}}\mathrm{x}\\ \mathrm{For}\text{n}=\text{1}\\ \text{y}=\text{sin x}\\ \text{Differentiating w.r.t. x, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{cosx}\\ \mathrm{For}\text{n}=\text{2}\\ \text{y}={\text{sin}}^{\text{2}}\text{x}\\ \text{Differentiating w.r.t. x, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{2}\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sinx}.\mathrm{sinx}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{sinx}\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}\mathrm{x}\right)+\mathrm{sin}\mathrm{x}\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}\mathrm{x}\right)\left[\mathrm{By}\text{Leibnitz product rule}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}+\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{sin}\mathrm{x}\mathrm{cos}\mathrm{x}...\left(\mathrm{i}\right)\\ \mathrm{For}\text{n}=\text{3}\\ \text{y}={\text{sin}}^{\text{3}}\text{x}\\ \text{Differentiating w.r.t. x, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{3}\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sinx}.{\mathrm{sin}}^{2}\mathrm{x}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{sin}\mathrm{x}\left(\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{2}\mathrm{x}\right)+{\mathrm{sin}}^{2}\mathrm{x}\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}\mathrm{x}\right)\left[\mathrm{By}\text{Leibnitz product rule}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{sin}\mathrm{x}\left(2\mathrm{sinxcosx}\right)+{\mathrm{sin}}^{2}\mathrm{x}\mathrm{cos}\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2{\mathrm{sin}}^{2}\mathrm{x}\mathrm{cos}\mathrm{x}+{\mathrm{sin}}^{2}\mathrm{x}\mathrm{cos}\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{sin}}^{2}\mathrm{x}\mathrm{cos}\mathrm{x}\\ \mathrm{According}\text{to this pattern, we can say}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{sin}}^{\mathrm{n}}\mathrm{x}\right)=\mathrm{n}{\mathrm{sin}}^{\left(\mathrm{n}-1\right)}\mathrm{x}\mathrm{cos}\mathrm{x}\\ \mathrm{Let}\text{this relation should be true for n}=\text{k}\\ \text{Then,\hspace{0.17em}}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{sin}}^{\mathrm{k}}\mathrm{x}\right)=\mathrm{k}{\mathrm{sin}}^{\left(\mathrm{k}-1\right)}\mathrm{x}\mathrm{cos}\mathrm{x}...\left(\mathrm{ii}\right)\\ \mathrm{Now},\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{sin}}^{\mathrm{k}+1}\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{sin}}^{\mathrm{k}}\mathrm{x}\mathrm{sin}\mathrm{x}\right)\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}={\mathrm{sin}}^{\mathrm{k}}\mathrm{x}\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\right)+\left(\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{\mathrm{k}}\mathrm{x}\right)\mathrm{sin}\mathrm{x}\\ ={\mathrm{sin}}^{\mathrm{k}}\mathrm{x}\left(\mathrm{cosx}\right)+\left(\mathrm{k}{\mathrm{sin}}^{\left(\mathrm{k}-1\right)}\mathrm{x}\mathrm{cos}\mathrm{x}\right)\mathrm{sin}\mathrm{x}\\ \left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\right]\\ ={\mathrm{sin}}^{\mathrm{k}}\mathrm{x}\mathrm{cos}\mathrm{x}+\mathrm{k}{\mathrm{sin}}^{\mathrm{k}}\mathrm{x}\mathrm{cos}\mathrm{x}\\ =\left(\mathrm{k}+1\right){\mathrm{sin}}^{\mathrm{k}}\mathrm{x}\mathrm{cos}\mathrm{x}\\ \mathrm{So},\text{the declaration of equation}\left(\mathrm{ii}\right)\text{is true for n}=\mathrm{k}+1.\\ \mathrm{Hence},\text{by Mathematical inducation,}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{sin}}^{\mathrm{n}}\mathrm{x}\right)=\mathrm{n}{\mathrm{sin}}^{\mathrm{n}-1}\mathrm{x}\mathrm{cos}\mathrm{x}\end{array}$

Q.50

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{a + bsinx}}}{\mathbf{\text{c + dcosx}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}+\mathrm{b}\mathrm{sin}\mathrm{x}}{\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{a}+\mathrm{b}\mathrm{sin}\mathrm{x}}{\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}}\right)\\ =\frac{\left\{\left(\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{a}+\mathrm{b}\mathrm{sin}\mathrm{x}\right)-\left(\mathrm{a}+\mathrm{b}\mathrm{sin}\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}\right)\right\}}{{\left(\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}\right)}^{2}}\\ =\frac{\left(\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}\right)\left(0+\mathrm{b}\mathrm{cos}\mathrm{x}\right)-\left(\mathrm{a}+\mathrm{b}\mathrm{sin}\mathrm{x}\right)\left(0-\mathrm{d}\mathrm{sin}\mathrm{x}\right)}{{\left(\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}\right)}^{2}}\\ =\frac{\mathrm{bc}\mathrm{cos}\mathrm{x}+\mathrm{bd}{\mathrm{cos}}^{2}\mathrm{x}+\mathrm{ad}\mathrm{sin}\mathrm{x}+\mathrm{bd}{\mathrm{sin}}^{2}\mathrm{x}}{{\left(\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}\right)}^{2}}\end{array}$ $\begin{array}{l}=\frac{\mathrm{bc}\mathrm{cos}\mathrm{x}+\mathrm{bd}\left({\mathrm{cos}}^{2}\mathrm{x}+{\mathrm{sin}}^{2}\mathrm{x}\right)+\mathrm{ad}\mathrm{sin}\mathrm{x}}{{\left(\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}\right)}^{2}}\\ =\frac{\mathrm{bc}\mathrm{cos}\mathrm{x}+\mathrm{bd}\left(1\right)+\mathrm{ad}\mathrm{sin}\mathrm{x}}{{\left(\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}\right)}^{2}}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{bc}\mathrm{cos}\mathrm{x}+\mathrm{ad}\mathrm{sin}\mathrm{x}+\mathrm{bd}}{{\left(\mathrm{c}+\mathrm{d}\mathrm{cos}\mathrm{x}\right)}^{2}}\end{array}$

Q.51

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{sin}}\left(\text{x+a}\right)}{\mathbf{\text{cosx}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\text{x}\right)=\frac{\mathrm{sin}\left(\text{x}+\text{a}\right)}{\mathrm{cosx}}\\ \text{}=\frac{\mathrm{sinx}\text{}\mathrm{cos}\text{a}+\mathrm{cosx}\text{}\mathrm{sin}\text{a}}{\mathrm{cos}\text{x}}\\ \mathrm{Differentiating}\text{w}.\text{r}.\text{t}.\text{x},\mathrm{we}\text{}\mathrm{get}\\ \left(\text{x}\right)=\frac{\text{d}}{\mathrm{dx}}\left(\frac{\mathrm{sin}\text{}\mathrm{x}\text{}\mathrm{cos}\text{a}+\mathrm{cos}\text{}\mathrm{x}\text{}\mathrm{sin}\text{a}}{\mathrm{cos}\text{x}}\right)\\ \text{}=\frac{\left\{\begin{array}{l}\mathrm{cos}\text{x}\frac{\text{d}}{\mathrm{dx}}\left(\mathrm{sin}\text{}\mathrm{x}\text{}\mathrm{cos}\text{a}+\mathrm{cos}\text{}\mathrm{x}\text{}\mathrm{sin}\text{a}\right)\\ \text{}-\left(\mathrm{sin}\text{}\mathrm{x}\text{}\mathrm{cos}\text{a}+\mathrm{cos}\text{}\mathrm{x}\text{}\mathrm{sin}\text{a}\right)\frac{\text{d}}{\mathrm{dx}}\mathrm{cos}\text{x}\end{array}\right\}}{{\left(\mathrm{cos}\text{x}\right)}^{2}}\\ \text{}=\frac{\left\{\begin{array}{l}\mathrm{cos}\text{x}\left(\mathrm{cos}\text{}\mathrm{x}\text{}\mathrm{cos}\text{a}-\mathrm{sin}\text{}\mathrm{x}\text{}\mathrm{sin}\text{a}\right)\\ \text{}-\left(\mathrm{sin}\text{}\mathrm{x}\text{}\mathrm{cos}\text{a}+\mathrm{cos}\text{}\mathrm{x}\text{}\mathrm{sin}\text{a}\right)\left(-\mathrm{sin}\text{x}\right)\end{array}\right\}}{{\mathrm{cos}}^{2}\text{x}}\\ \text{}=\frac{\left\{{\mathrm{cos}}^{2}\text{}\mathrm{x}\text{}\mathrm{cos}\text{a}-\mathrm{sin}\text{}\mathrm{x}\text{}\mathrm{cos}\text{}\mathrm{x}\text{}\mathrm{sin}\text{a}+{\mathrm{sin}}^{2}\text{}\mathrm{x}\text{}\mathrm{cos}\text{a}+\mathrm{sin}\text{}\mathrm{x}\text{}\mathrm{cos}\text{}\mathrm{x}\text{}\mathrm{sin}\text{a}\right\}}{{\mathrm{cos}}^{2}\text{x}}\\ \text{}=\frac{\left({\mathrm{cos}}^{2}\mathrm{xcos}\text{a}+{\mathrm{sin}}^{2}\mathrm{xcos}\text{a}\right)}{{\mathrm{cos}}^{2}\text{x}}\\ \text{}=\frac{\left({\mathrm{cos}}^{2}\text{x}+{\mathrm{sin}}^{2}\text{x}\right)\mathrm{cos}\text{a}}{{\mathrm{cos}}^{2}\text{x}}\\ \text{}=\frac{1×\mathrm{cos}\text{a}}{{\mathrm{cos}}^{2}\text{x}}\\ \text{}\left(\text{x}\right)=\frac{\mathrm{cos}\text{a}}{{\mathrm{cos}}^{2}\text{x}}\end{array}$

Q.52 Find the derivative of the function x4 (5sin x – 3cos x).

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)={\mathrm{x}}^{4}\left(5\mathrm{sinx}-3\mathrm{cosx}\right)\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\end{array}$ $\begin{array}{l}\text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\mathrm{x}}^{4}\left(5\mathrm{sinx}-3\mathrm{cosx}\right)\right\}\\ \mathrm{By}\text{Leibnitz product rule:}\\ \text{f’}\left(\mathrm{x}\right)={\mathrm{x}}^{4}\frac{\mathrm{d}}{\mathrm{dx}}\left(5\mathrm{sinx}-3\mathrm{cosx}\right)+\left(5\mathrm{sinx}-3\mathrm{cosx}\right)\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{4}\\ ={\mathrm{x}}^{4}\left(5\mathrm{cosx}+3\mathrm{sinx}\right)+\left(5\mathrm{sinx}-3\mathrm{cosx}\right)×4{\mathrm{x}}^{3}\\ ={\mathrm{x}}^{3}\left(5\mathrm{xcosx}+3\mathrm{xsinx}+20\mathrm{sinx}-12\mathrm{cosx}\right)\end{array}$

Q.53 Find the derivative of the function (x2 + 1) cosx.

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\left({\mathrm{x}}^{2}+1\right)\mathrm{cosx}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left({\mathrm{x}}^{2}+1\right)\mathrm{cosx}\right\}\\ \mathrm{By}\text{Leibnitz product rule:}\end{array}$ $\begin{array}{l}\text{f’}\left(\mathrm{x}\right)=\left({\mathrm{x}}^{2}+1\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}+\mathrm{cosx}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}+1\right)\\ =\left({\mathrm{x}}^{2}+1\right)\left(-\mathrm{sinx}\right)+\mathrm{cosx}\left(2\mathrm{x}+0\right)\\ =-{\mathrm{x}}^{2}\mathrm{sinx}-\mathrm{sinx}+2\mathrm{xcosx}\end{array}$

Q.54 Find the derivative of the function (ax2 + sin x)(p + q cos x).

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\left({\mathrm{ax}}^{2}+\mathrm{sinx}\right)\left(\mathrm{p}+\mathrm{qcosx}\right)\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left({\mathrm{ax}}^{2}+\mathrm{sinx}\right)\left(\mathrm{p}+\mathrm{qcosx}\right)\right\}\\ \mathrm{By}\text{Leibnitz product rule:}\end{array}$ $\text{f’}\left(\mathrm{x}\right)=\left\{\left({\mathrm{ax}}^{2}+\mathrm{sinx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{p}+\mathrm{qcosx}\right)+\left(\mathrm{p}+\mathrm{qcosx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{ax}}^{2}+\mathrm{sinx}\right)\right\}$ $\begin{array}{l}\text{}\text{}\text{}\end{array}$ $\begin{array}{l}=\left\{\begin{array}{l}\left({\mathrm{ax}}^{2}+\mathrm{sinx}\right)\left(0-\mathrm{qsinx}\right)+\\ \left(\mathrm{p}+\mathrm{q}\mathrm{cosx}\right)\left(2\mathrm{ax}+\mathrm{cosx}\right)\end{array}\right\}\\ =-\text{\hspace{0.17em}}\mathrm{q}\mathrm{sin}\mathrm{x}\left({\mathrm{ax}}^{2}+\mathrm{sinx}\right)+\left(\mathrm{p}+\mathrm{q}\mathrm{cos}\mathrm{x}\right)\left(2\mathrm{ax}+\mathrm{cosx}\right)\end{array}$

Q.55 Find the derivative of the function (x + cos x)(x – tan x).

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{cosx}\right)\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left(\mathrm{x}+\mathrm{cosx}\right)\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)\right\}\\ \text{f’}\left(\mathrm{x}\right)=\left\{\left(\mathrm{x}+\mathrm{cosx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)+\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\mathrm{cosx}\right)\right\}\end{array}$ $\mathrm{By}\text{Leibnitz product rule:}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}\text{f’}\left(\mathrm{x}\right)=\left\{\left(\mathrm{x}+\mathrm{cosx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)+\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\mathrm{cosx}\right)\right\}\\ =\left(\mathrm{x}+\mathrm{cosx}\right)\left(1-{\mathrm{sec}}^{2}\mathrm{x}\right)+\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)\left(1-\mathrm{sinx}\right)\\ =-\text{\hspace{0.17em}}\left(\mathrm{x}+\mathrm{cosx}\right)\left({\mathrm{sec}}^{2}\mathrm{x}-1\right)+\left(\mathrm{x}-\mathrm{tanx}\right)\left(1-\mathrm{sinx}\right)\\ =-\text{\hspace{0.17em}}{\mathrm{tan}}^{2}\mathrm{x}\left(\mathrm{x}+\mathrm{cosx}\right)+\left(\mathrm{x}-\mathrm{tanx}\right)\left(1-\mathrm{sinx}\right)\end{array}$

Q.56

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{4x+5sinx}}}{\mathbf{\text{3x+7cosx}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}+5\mathrm{sinx}}{3\mathrm{x}+7\mathrm{cosx}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{4\mathrm{x}+5\mathrm{sinx}}{3\mathrm{x}+7\mathrm{cosx}}\right)\end{array}$ $\begin{array}{l}=\frac{\left\{\begin{array}{l}\left(3\mathrm{x}+7\mathrm{cosx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(4\mathrm{x}+5\mathrm{sinx}\right)\\ -\left(4\mathrm{x}+5\mathrm{sinx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(3\mathrm{x}+7\mathrm{cosx}\right)\end{array}\right\}}{{\left(3\mathrm{x}+7\mathrm{cosx}\right)}^{2}}\\ =\frac{\left(3\mathrm{x}+7\mathrm{cosx}\right)\left(4+5\mathrm{cosx}\right)-\left(4\mathrm{x}+5\mathrm{sinx}\right)\left(3-7\mathrm{sinx}\right)}{{\left(3\mathrm{x}+7\mathrm{cosx}\right)}^{2}}\\ =\frac{\left(\begin{array}{l}12\mathrm{x}+28\mathrm{cosx}+15\mathrm{xcosx}+35{\mathrm{cos}}^{2}\mathrm{x}-12\mathrm{x}-15\mathrm{sinx}\\ +28\mathrm{xsinx}+35{\mathrm{sin}}^{2}\mathrm{x}\end{array}\right)}{{\left(3\mathrm{x}+7\mathrm{cosx}\right)}^{2}}\\ =\frac{\left(\begin{array}{l}28\mathrm{cosx}+15\mathrm{xcosx}+35{\mathrm{cos}}^{2}\mathrm{x}-15\mathrm{sinx}\\ +28\mathrm{xsinx}+35\left(1-{\mathrm{cos}}^{2}\mathrm{x}\right)\end{array}\right)}{{\left(3\mathrm{x}+7\mathrm{cosx}\right)}^{2}}\end{array}$ $\begin{array}{l}\text{}\text{}\end{array}$ $\begin{array}{l}=\frac{\left(\begin{array}{l}28\mathrm{cosx}+15\mathrm{xcosx}+35{\mathrm{cos}}^{2}\mathrm{x}-15\mathrm{sinx}\\ +28\mathrm{xsinx}+35-35{\mathrm{cos}}^{2}\mathrm{x}\end{array}\right)}{{\left(3\mathrm{x}+7\mathrm{cosx}\right)}^{2}}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\left(35+15\mathrm{xcosx}+28\mathrm{cosx}+28\mathrm{xsinx}-15\mathrm{sinx}\right)}{{\left(3\mathrm{x}+7\mathrm{cosx}\right)}^{2}}\end{array}$

Q.57

$\mathbf{\text{Find the derivative of the function}}\frac{{\mathbf{\text{x}}}^{\mathbf{\text{2}}}\mathbf{\text{\hspace{0.17em}cos}}\left(\frac{\pi }{\text{4}}\right)}{\mathbf{\text{sinx}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)}{\mathrm{sinx}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\end{array}$ $\begin{array}{l}\text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)}{\mathrm{sinx}}\right)\\ =\frac{\mathrm{sinx}\left\{\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)\right\}-{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\right)}{{\left(\mathrm{sinx}\right)}^{2}}\\ =\frac{\mathrm{sinx}\left\{2\mathrm{x}\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)\right\}-{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)\left(\mathrm{cosx}\right)}{{\mathrm{sin}}^{2}\mathrm{x}}\\ =\frac{2\mathrm{xsinxcos}\left(\frac{\mathrm{\pi }}{4}\right)-{\mathrm{x}}^{2}\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)\left(\mathrm{cosx}\right)}{{\mathrm{sin}}^{2}\mathrm{x}}\\ =\frac{\mathrm{xcos}\left(\frac{\mathrm{\pi }}{4}\right)\left(2\mathrm{sinx}-\mathrm{x}\text{\hspace{0.17em}}\mathrm{cosx}\right)}{{\mathrm{sin}}^{2}\mathrm{x}}\end{array}$

Q.58

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{x}}}{\mathbf{\text{1+tanx}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{1+\mathrm{tanx}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{1+\mathrm{tanx}}\right)\\ =\frac{\left(1+\mathrm{tanx}\right)\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)-\mathrm{x}\text{\hspace{0.17em}}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+\mathrm{tanx}\right)}{{\left(1+\mathrm{tanx}\right)}^{2}}\\ =\frac{\left(1+\mathrm{tanx}\right)\left(1\right)-\mathrm{x}\text{\hspace{0.17em}}\left(0+{\mathrm{sec}}^{2}\mathrm{x}\right)}{{\left(1+\mathrm{tanx}\right)}^{2}}\\ =\frac{1+\mathrm{tanx}-\mathrm{x}\text{\hspace{0.17em}}{\mathrm{sec}}^{2}\mathrm{x}}{{\left(1+\mathrm{tanx}\right)}^{2}}\end{array}$

Q.59 Find the derivative of the function (x + sec x)(x – tan x).

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{secx}\right)\left(\mathrm{x}-\mathrm{tanx}\right)\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left(\mathrm{x}+\mathrm{secx}\right)\left(\mathrm{x}-\mathrm{tanx}\right)\right\}\\ \mathrm{By}\text{Leibnitz product rule:}\\ \text{f’}\left(\mathrm{x}\right)=\left\{\left(\mathrm{x}+\mathrm{secx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)+\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\mathrm{secx}\right)\right\}\end{array}$ $\text{f’}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{secx}\right)\left(1-{\mathrm{sec}}^{2}\mathrm{x}\right)+\left(\mathrm{x}-\mathrm{tan}\text{}\mathrm{x}\right)\left(1+\mathrm{sec}\mathrm{x}\mathrm{tan}\mathrm{x}\right)$

Q.60

$\mathbf{\text{Find the derivative of the function}}\frac{\mathbf{\text{x}}}{{\mathbf{\text{sin}}}^{\mathbf{\text{n}}}\mathbf{\text{x}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\mathrm{sin}}^{\mathrm{n}}\mathrm{x}}\\ \mathrm{Differentiating}\text{w.r.t. x, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}f’}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{{\mathrm{sin}}^{\mathrm{n}}\mathrm{x}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{sin}}^{\mathrm{n}}\mathrm{x}\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)-\mathrm{x}\text{\hspace{0.17em}}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{\mathrm{n}}\mathrm{x}}{{\left({\mathrm{sin}}^{\mathrm{n}}\mathrm{x}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{sin}}^{\mathrm{n}}\mathrm{x}\left(1\right)-\mathrm{x}\text{\hspace{0.17em}}\left({\mathrm{nsin}}^{\mathrm{n}-1}\mathrm{xcosx}\right)}{{\left({\mathrm{sin}}^{\mathrm{n}}\mathrm{x}\right)}^{2}}\\ \left[âˆµ\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{\mathrm{n}}\mathrm{x}={\mathrm{nsin}}^{\mathrm{n}-1}\mathrm{xcosx}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{sin}}^{\mathrm{n}-1}\mathrm{x}\left(\mathrm{sinx}-\mathrm{nx}\text{\hspace{0.17em}}\mathrm{cosx}\right)}{{\mathrm{sin}}^{2\mathrm{n}}\mathrm{x}}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{{\mathrm{sin}}^{\mathrm{n}}\mathrm{x}}\right)=\frac{\mathrm{sinx}-\mathrm{nx}\text{\hspace{0.17em}}\mathrm{cosx}}{{\mathrm{sin}}^{\mathrm{n}+1}\mathrm{x}}\end{array}$

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### 1. What are the topics covered in the NCERT Solutions for Class 11 Mathematics Chapter 13?

The NCERT Solutions for Class 11 Mathematics Chapter 13 has complete information about the limits and derivatives. The topics like intuitive ideas of derivatives, limits, algebra of limits, limits of trigonometric functions, derivatives, algebra of derivatives of functions, derivatives of polynomials and trigonometric functions are explained with examples. The students will get detailed and authentic solutions without having to look anywhere else.

### 2. What are the important points should I keep in mind while solving problems in Mathematics?

Students should keep in mind the following important points while solving problems in Mathematics:

1. They must read each question carefully and attentively. Analyse the type of problem.
2. They must note down all the conditions given in the questions on a piece of paper.
3. They must solve it in a step-by-step format.
4. They must begin with easy steps and calculations.
5. They must double-check from the start once the calculation is done.