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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

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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.

$\lim_{x \to 3} (x + 3)$

$\lim_{x \to π} (x - \frac{22}{7})$

$\lim_{r \to 1} {πr}^{2}$

$\lim_{x \to 4} (\frac{4 x + 3}{x - 2})$

$\lim_{x \to - 1} (\frac{x^{10} + x^{5} + 1}{x - 1})$

$\lim_{x \to 0} \frac{{(x + 1)}^{5} - 1}{x}$

$\lim_{x \to 2} (\frac{3 x^{2} - x - 10}{x^{2} - 4})$

$\lim_{x \to 3} \frac{x^{4} - 81}{2 x^{2} - 5 x - 3}$

$\lim_{x \to 0} (\frac{ax + b}{cx + 1})$

10.

$\lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$

11.

$\lim_{x \to 1} (\frac{{ax}^{2} + bx + c}{{cx}^{2} + bx + a}), a + b + c \neq 0$

12.

$\lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$

13.

$\lim_{x \to 0} \frac{sinax}{bx}$

14.

$\lim_{x \to 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$

15.

$\lim_{x \to π} \frac{\sin (π - x)}{π (π - x)}$

16.

$\lim_{x \to 0} \frac{\cos x}{(π - x)}$

17.

$\lim_{x \to 0} \frac{\cos 2 x - 1}{cosx - 1}$

18.

$\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}$

19.

$\lim_{x \to 0} x \sec x$

20.

$\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}, a, b, a + b \neq 0$

21.

$\lim_{x \to 0} (cosec x - \cot x)$

22.

$\lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}}$

Ans

$\begin{array}{l} \lim_{x \to 3} (x + 3) = 3 + 3 \\ = 6 \end{array}$

$\lim_{x \to π} (x - \frac{22}{7}) = π - \frac{22}{7}$

$\begin{array}{l} \lim_{r \to 1} {πr}^{2} = π {(1)}^{2} \\ = π \end{array}$

$\begin{array}{l} \lim_{x \to 4} (\frac{4 x + 3}{x - 2}) = \frac{4 (4) + 3}{4 - 2} \\ = \frac{19}{2} \end{array}$

$\begin{array}{l} \lim_{x \to - 1} (\frac{x^{10} + x^{5} + 1}{x - 1}) = \frac{{(- 1)}^{10} + {(- 1)}^{5} + 1}{(- 1) - 1} \\ = \frac{1 - 1 + 1}{- 1 - 1} \\ = - \frac{1}{2} \end{array}$

$\begin{array}{l} \lim_{x \to 0} \frac{{(x + 1)}^{5} - 1}{x} \\ Let x + 1 = y \Rightarrow x = y - 1 \\ Then, \\ \lim_{y \to 1} \frac{y^{5} - 1}{y - 1} = \lim_{y \to 1} \frac{y^{5} - 1^{5}}{y - 1} \\ = 5 {(1)}^{5 - 1} [âˆµ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = {na}^{n - 1}] \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = 5 \\ ∴ \lim_{x \to 0} \frac{{(x + 1)}^{5} - 1}{x} = 5 . \end{array}$

$\begin{array}{l} \lim_{x \to 2} (\frac{3 x^{2} - x - 10}{x^{2} - 4}) = \lim_{x \to 2} {\frac{3 x^{2} - 6 x + 5 x - 10}{(x - 2) (x + 2)}} \\ = \lim_{x \to 2} {\frac{3 x (x - 2) + 5 (x - 2)}{(x - 2) (x + 2)}} \\ = \lim_{x \to 2} {\frac{(x - 2) (3 x + 5)}{(x - 2) (x + 2)}} \\ = \lim_{x \to 2} {\frac{(3 x + 5)}{(x + 2)}} \\ = \frac{3 (2) + 5}{2 + 2} \\ = \frac{11}{4} \end{array}$

$\begin{array}{l} \lim_{x \to 3} \frac{x^{4} - 81}{2 x^{2} - 5 x - 3} = \lim_{x \to 3} \frac{{(x^{2})}^{2} - 9^{2}}{2 x^{2} - 6 x + x - 3} \\ = \lim_{x \to 3} \frac{(x^{2} - 9) (x^{2} + 9)}{2 x (x - 3) + 1 (x - 3)} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = \lim_{x \to 3} \frac{(x - 3) (x + 3) (x^{2} + 9)}{(x - 3) (2 x + 1)} \\ = \lim_{x \to 3} \frac{(x + 3) (x^{2} + 9)}{(2 x + 1)} \\ = \lim_{x \to 3} \frac{(3 + 3) (3^{2} + 9)}{(2 \times 3 + 1)} \\ = \frac{6 \times 18}{7} \\ = \frac{108}{7} \end{array}$

$\begin{array}{l} \lim_{x \to 0} (\frac{ax + b}{cx + 1}) = \frac{a (0) + b}{c (0) + 1} \\ = \frac{b}{1} \\ = b \end{array}$

10.

$\begin{array}{l} \lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1} = \lim_{z \to 1} \frac{{(z^{\frac{1}{6}})}^{2} - 1}{z^{\frac{1}{6}} - 1} \\ = 2 (1^{2 - 1}) [âˆµ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = {na}^{n - 1}] \\ = 2 \end{array}$

11.

$\begin{array}{l} \lim_{x \to 1} (\frac{{ax}^{2} + bx + c}{{cx}^{2} + bx + a}) = \frac{a {(1)}^{2} + b (1) + c}{c {(1)}^{2} + b (1) + a} \\ = \frac{a + b + c}{c + b + a} \\ = 1 \end{array}$

12.

$\begin{array}{l} \lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \lim_{x \to - 2} \frac{\frac{2 + x}{2 x}}{x + 2} \\ = \lim_{x \to - 2} \frac{x + 2}{2 x (x + 2)} \\ \lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \frac{1}{2 (- 2)} \\ = - \frac{1}{4} \end{array}$

13.

$\begin{array}{l} \lim_{x \to 0} \frac{\sin ax}{bx} = \frac{1}{b} \lim_{x \to 0} \frac{\sin ax}{ax} \times a \\ = \frac{a}{b} \lim_{ax \to 0} (\frac{\sin ax}{ax}) \\ = \frac{a}{b} \times 1 [âˆµ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = \frac{a}{b} \end{array}$

14.

$\begin{array}{l} \lim_{x \to 0} \frac{sinax}{sinbx} = \frac{a \lim_{ax \to 0} (\frac{sinax}{ax})}{b \lim_{bx \to 0} (\frac{sinbx}{bx})} \\ = \frac{a \times 1}{b \times 1} [âˆµ \lim_{x \to 0} \frac{sinx}{x} = 1] \\ = \frac{a}{b} \end{array}$

15.

$\begin{array}{l} \lim_{x \to π} \frac{\sin (π - x)}{π (π - x)} = \frac{1}{π} {\lim_{π - x \to 0} \frac{\sin (π - x)}{(π - x)}} \\ = \frac{1}{π} (1) \\ = \frac{1}{π} \end{array}$

16.

$\begin{array}{l} \lim_{x \to 0} \frac{cosx}{(π - x)} = \frac{\cos 0}{(π - 0)} \\ = \frac{1}{π} [âˆµ \cos 0 = 1] \end{array}$

17.

$\begin{array}{l} \lim_{x \to 0} \frac{\cos 2 x - 1}{cosx - 1} = \lim_{x \to 0} \frac{1 - 2 \sin^{2} x - 1}{1 - 2 \sin^{2} (\frac{x}{2}) - 1} \\ = \lim_{x \to 0} \frac{- 2 \sin^{2} x}{- 2 \sin^{2} (\frac{x}{2})} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = \lim_{x \to 0} \frac{(\frac{\sin^{2} x}{x^{2}})}{{\frac{\sin^{2} (\frac{x}{2})}{4 {(\frac{x}{2})}^{2}}}} \\ = 4 \lim_{x \to 0} \frac{{(\frac{sinx}{x})}^{2}}{{\frac{\sin (\frac{x}{2})}{(\frac{x}{2})}}^{2}} \\ = 4 \frac{{(1)}^{2}}{{(1)}^{2}} \\ \lim_{x \to 0} \frac{\cos 2 x - 1}{cosx - 1} = 4 \end{array}$

18.

$\begin{array}{l} \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} = \lim_{x \to 0} \frac{x (a + cosx)}{b \sin x} \\ = \frac{\lim_{x \to 0} (a + \cos x)}{b \lim_{x \to 0} (\frac{\sin x}{x})} \\ = \frac{(a + \cos 0)}{b (1)} \\ \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} = \frac{a + 1}{b} \end{array}$

19.

$\begin{array}{l} \lim_{x \to 0} x \sec x = 0 . \sec 0 \\ = 0 \times 1 \\ = 0 \end{array}$

20.

$\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} = \lim_{x \to 0} \frac{x (\frac{a \sin ax}{ax} + b)}{x (a + \frac{b \sin bx}{bx})}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = \lim_{x \to 0} \frac{(\frac{a \sin ax}{ax} + b)}{(a + \frac{b \sin bx}{bx})} \\ = \frac{(a \times 1 + b)}{(a + b \times 1)} \\ = \frac{a + b}{a + b} \\ \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} = 1 \end{array}$

21.

$\begin{array}{l} \lim_{x \to 0} (cosec x - \cot x) = \lim_{x \to 0} (\frac{1}{sinx} - \frac{\cos x}{sinx}) \\ = \lim_{x \to 0} (\frac{1 - \cos x}{sinx}) \\ = \lim_{x \to 0} (\frac{2 \sin^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}) \\ = \lim_{x \to 0} (\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}) \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = \lim_{x \to 0} (\tan \frac{x}{2}) \\ = \tan 0 \\ \lim_{x \to 0} (cosec x - \cot x) = 0 \end{array}$

22.

$\begin{array}{l} Given : \lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} \\ Let x - \frac{π}{2} = y \Rightarrow x = \frac{π}{2} + y \\ If x \to \frac{π}{2} then y \to 0 \\ ∴ \lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} = \lim_{y \to 0} \frac{\tan 2 (\frac{π}{2} + y)}{\frac{π}{2} + y - \frac{π}{2}} \\ = \lim_{y \to 0} \frac{\tan (π + 2 y)}{y} \\ = \lim_{y \to 0} \frac{\tan 2 y}{y} [âˆµ \tan (π + x) = \tan x] \\ = 2 \lim_{y \to 0} \frac{\tan 2 y}{2 y} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = 2 \times 1 [\lim_{x \to 0} \frac{\tan x}{x} = 1] \\ \lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} = 2 \end{array}$

Q.2

$Find \lim_{x \to 0} f (x) and \lim_{x \to 1} f (x), wheref (x) = {\begin{cases} 2 x + 3, x \leq 0 \\ 3 (x + 1), x > 0 \end{cases}$

Ans

$\begin{array}{l} The given function is : f (x) = {\begin{cases} 2 x + 3, x \leq 0 \\ 3 (x + 1), x > 0 \end{cases} \\ At x = 0, \\ L . H . L . = \lim_{x \to 0^{-}} f (x) \\ = \lim_{x \to 0} (2 x + 3) \\ = 2 (0) + 3 \\ = 3 \\ R . H . L . = \lim_{x \to 0^{+}} f (x) \\ = \lim_{x \to 0} 3 (x + 1) \\ = 3 (0 + 1) \\ = 3 \\ ∴ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} f (x) = 3 \\ At x = 1, \\ L . H . L . = \lim_{x \to 1^{-}} f (x) \end{array}$ $\begin{array}{l} = \lim_{x \to 1} 3 (x + 1) \\ = 3 (1 + 1) \\ = 6 \\ R . H . L . = \lim_{x \to 1^{+}} f (x) \\ = \lim_{x \to 1} 3 (x + 1) \\ = 3 (1 + 1) \\ = 6 \\ ∴ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1} f (x) = 6 \end{array}$

Q.3

$Find \lim_{x \to 1} f (x), where f (x) = {\begin{cases} x^{2} - 1, x \leq 1 \\ - x^{2} - 1, x > 1 \end{cases}$

Ans

$\begin{array}{l} The given function is; f (x) = {\begin{cases} x^{2} - 1, x \leq 1 \\ - x^{2} - 1, x > 1 \end{cases} \\ At x = 1, \\ L . H . L . = \lim_{x \to 1^{-}} f (x) \\ = \underset{x \to 1}{im} (x^{2} - 1) \\ = {(1)}^{2} - 1 \\ = 0 \\ R . H . L . = \lim_{x \to 1^{+}} f (x) \\ = \lim_{x \to 1} (- x^{2} - 1) \\ = - {(1)}^{2} - 1 \\ = - 2 \\ ∴ \lim_{x \to 1^{-}} f (x) \lim_{x \to 1^{+}} f (x) \\ Thus, \lim_{x \to 1} f (x) does not exist . \end{array}$

Q.4

$Find \lim_{x \to 0} f (x), where f (x) = {\begin{cases} \frac{x}{| x |}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

Ans

$\begin{array}{l} The given function is, \\ f (x) = {\begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 0, & x = 0 \end{cases} \\ Atx = 0, \\ L . H . L . = lim_{x \to 0^{-}} f (x) \\ = lim_{x \to 0^{-}} (\frac{x}{|x|}) \\ = lim_{x \to 0} (\frac{x}{- x}) [whenx < 0, |x| = - x] \\ = - 1 \\ R . H . L . = lim_{x \to 0^{+}} f (x) \\ = lim_{x \to 0^{+}} (\frac{x}{|x|}) \\ = lim_{x \to 0} (\frac{x}{x}) [Whenx > 0, |x| = x] \\ = 1 \\ Since, lim_{x \to 0^{-}} f (x) lim_{x \to 0^{+}} f (x) \\ So, lim_{x \to 0} f (x) doesnotexist . \end{array}$

Q.5

$\begin{array}{l} Suppose f (x) = {\begin{cases} a + bx, x < 1 \\ 4, x = 1 \\ b - ax, x > 1 \end{cases} \\ and if \lim_{x \to 1} f (x) = f (1) what are possible values of a and b ? \end{array}$

Ans

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

Q.6

$Find \lim_{x \to 5} f (x), where f (x) = | x | - 5$

Ans

$\begin{array}{l} The given function is, \\ f (x) = | x | - 5 = {\begin{cases} - x - 5, x < 0 \\ - 5, x = 0 \\ x - 5, x > 0 \end{cases} \\ For x = 5, \\ L . H . L . = \lim_{x \to 5^{-}} f (x) \\ = \lim_{x \to 5} (x - 5) [When x>0, | x | = x] \end{array}$ $\begin{array}{l} = 5 - 5 \\ = 0 \\ R . H . L . = \lim_{x \to 5^{+}} f (x) \\ = \lim_{x \to 5} (x - 5) [When x>0, | x | = x] \\ = 5 - 5 \\ = 0 \\ ∴ \lim_{x \to 5^{-}} f (x) = \lim_{x \to 5^{+}} f (x) = \lim_{x \to 5} f (x) = 0 \\ Thus, \lim_{x \to 5} f (x) = 0 \end{array}$

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

$What is \lim_{x \to a_{1}} f (x) ? For some a \neq a_{1}, a_{2}, . .., a_{n}, compute \lim_{x \to a} f (x) .$

Ans

$\begin{array}{l} The given function is: \\ f (x) = (x - a_{1}) (x - a_{2}) \dots (x - a_{n}) \\ ∴ \lim_{x \to a_{1}} f (x) = \lim_{x \to a_{1}} {(x - a_{1}) (x - a_{2}) \dots (x - a_{n})} \\ = {\lim_{x \to a_{1}} (x - a_{1})} {\lim_{x \to a_{1}} (x - a_{2})} \dots {\lim_{x \to a_{1}} (x - a_{n})} \\ = (a_{1} - a_{1}) (a_{1} - a_{2}) \dots (a_{1} - a_{n}) \\ = (0) (a_{1} - a_{2}) \dots (a_{1} - a_{n}) \\ = 0 \end{array}$ $And \lim_{x \to a} f (x) = \lim_{x \to a} {(x - a_{1}) (x - a_{2}) \dots (x - a_{n})}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = {\lim_{x \to a} (x - a_{1})} {\lim_{x \to a} (x - a_{2})} \dots {\lim_{x \to a} (x - a_{n})} \\ = (a - a_{1}) (a - a_{2}) \dots (a - a_{n}) \end{array}$

Q.8

$\begin{array}{l} If f (x) = {\begin{cases} | x | + 1, & x < 0 \\ 0, & x = 0 \\ | x | - 1, & x > 0 \end{cases} \\ For what value (s) of a does \lim_{x \to a} f (x) exists ? \end{array}$

Ans

$\begin{array}{l} The given function is: \\ f (x) = {\begin{cases} | x | + 1, x < 0 \\ 0, x = 0 \\ | x | - 1, x > 0 \end{cases} \\ For existence of \lim_{x \to a} f (x), \end{array}$ $\begin{array}{l} CaseI:Whena=0 \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (|x| + 1) \\ = \lim_{x \to 0} (- x + 1) [When x < 0, |x| = - x] \\ = 0 + 1 \\ = 1 \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (|x| - 1) \\ = \lim_{x \to 0} (x - 1) [When x > 0, |x| = x] \\ = 0 - 1 \\ = - 1 \\ Here, \lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x) \\ ∴ \lim_{x \to 0} f (x) does not exist. \\ Case II : When a < 0 \\ \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{-}} (|x| + 1) \\ = \lim_{x \to a} (- x + 1) [When x < 0, |x| = - x] \\ = - a + 1 \\ \lim_{x \to a^{+}} f (x) = \lim_{x \to a^{+}} (|x| + 1) \\ = \lim_{x \to a} (- x + 1) [When a < x < 0, |x| = - x] \\ = - a + 1 \\ ∴ \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) = - a + 1 \end{array}$ $\begin{array}{l} Thus, \lim_{x \to a} f (x) exists at x = a, where a<0. \\ Case III : When a > 0 \\ \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{-}} (| x | - 1) \\ = \lim_{x \to a} (x - 1) [When 0 < x < a, | x | = x] \\ = a - 1 \\ \lim_{x \to a^{+}} f (x) = \lim_{x \to a^{+}} (| x | - 1) \\ = \lim_{x \to a} (x - 1) [When a < x < 0, | x | = x] \\ = a - 1 \\ ∴ \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) = a - 1 \\ Thus, \lim_{x \to a} f (x) exists at x = a, where a>0. \\ Therefore, \lim_{x \to a} f (x) exists for all a \neq 0. \end{array}$

Q.9

$\begin{array}{l} If the function f (x) satisfies \lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π, \\ evaluate \lim_{x \to 1} f (x) . \end{array}$

Ans

$\begin{array}{l} G iven: \lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π \\ \lim_{x \to 1} {f (x) - 2} = \lim_{x \to 1} π (x^{2} - 1) \\ \Rightarrow \lim_{x \to 1} f (x) - \lim_{x \to 1} 2 = π (1^{2} - 1) \\ \Rightarrow \lim_{x \to 1} f (x) - \lim_{x \to 1} 2 = π (0) \end{array}$ $\begin{array}{l} \Rightarrow \lim_{x \to 1} f (x) = \lim_{x \to 1} 2 \\ \Rightarrow \lim_{x \to 1} f (x) = 2 . \end{array}$

Q.10

$\begin{array}{l} If f (x) = {\begin{cases} {mx}^{2} + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ {nx}^{3} + m, & x > 0 \end{cases} . \\ For what integers m and n does both \lim_{x \to 0} f (x) and \lim_{x \to 1} f (x) exist ? \end{array}$

Ans

$\begin{array}{l} We have, \\ f (x) = {\begin{cases} {mx}^{2} + n, x < 0 \\ nx + m, 0 \leq x \leq 1 \\ {nx}^{3} + m, x > 0 \end{cases} \\ For x = 0, \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0} ({mx}^{2} + n) \\ = n \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} ({nx}^{3} + m) \\ = m \\ Since, \lim_{x \to 0} f (x) exists. So, \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) \\ \Rightarrow n = m \\ Therefore, \lim_{x \to 0} f (x) exists if m = n. \\ For x = 1, \end{array}$ $\begin{array}{l} \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1} (nx + m) \\ = n + m \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1} ({nx}^{3} + m) \\ = n + m \\ Since, \lim_{x \to 0} f (x) exists. So, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1} f (x) \\ Therefore, \lim_{x \to 1} f (x) exists for any value of m and n. \end{array}$

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

Ans

$\begin{array}{l} We are given: \\ f (x) = x^{2} - 2 \\ Since, f ‘ (10) = \lim_{h \to 0} \frac{f (10 + h) - f (10)}{h} \\ ∴ f ‘ (10) = \lim_{h \to 0} \frac{{{(10 + h)}^{2} - 2} - {{(10)}^{2} - 2}}{h} \\ = \lim_{h \to 0} \frac{(100 + 20 h + h^{2} - 2) - (100 - 2)}{h} \\ = \lim_{h \to 0} \frac{98 + 20 h + h^{2} - 98}{h} \end{array}$ $\begin{array}{l} = \lim_{h \to 0} \frac{h (20 + h)}{h} \\ f ‘ (10) = 20 \\ Thus, {the derivative of the function x}^{2} - 2 at x = 10 is 20. \end{array}$

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

Ans

$\begin{array}{l} Wearegiven:f (x) =99x \\ Since, f ‘ (100) = \lim_{h \to 0} \frac{f (100 + h) - f (100)}{h} \\ ∴ f ‘ (100) = \lim_{h \to 0} \frac{99 (100 + h) - 99 (100)}{h} \\ = \lim_{h \to 0} \frac{9900 + 99 h - 9900}{h} \\ = \lim_{h \to 0} \frac{99 h}{h} \\ f ‘ (100) = 99 \\ Thus,thederivativeofthefunction99xatx=100is99. \end{array}$

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

Ans

$\begin{array}{l} Wearegiven: \\ f (x) = x \\ Since, f ‘ (1) = \underset{}{\lim_{h \to 0}} \frac{f (1 + h) - f (1)}{h} \\ ∴ f ‘ (1) = \lim_{h \to 0} \frac{(1 + h) - (1)}{h} \\ = \lim_{h \to 0} \frac{1 + h - 1}{h} \\ = \lim_{h \to 0} \frac{h}{h} \\ f ‘ (1) = 1 \\ Thus,thederivativeofthefunctionxatx=1is1. \end{array}$

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

$\begin{array}{l} (i) x^{3} - 27 (ii) (x - 1) (x - 2) \\ (iii) \frac{1}{x^{2}} (iv) \frac{x + 1}{x - 1} \end{array}$

Ans

$\begin{array}{l} Since, \\ f ‘ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ (i) f (x) = x^{3} - 27 \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{{{(x + h)}^{3} - 27} - (x^{3} - 27)}{h} \\ = \lim_{h \to 0} \frac{x^{3} + 3 x^{2} h + 3 {xh}^{2} + h^{3} - 27 - x^{3} + 27}{h} \\ = \lim_{h \to 0} \frac{3 x^{2} h + 3 {xh}^{2} + h^{3}}{h} \\ = \lim_{h \to 0} \frac{(3 x^{2} + 3 xh + h^{2}) h}{h} \\ = 3 x^{2} + 3 x (0) + {(0)}^{2} \\ f ‘ (x) = 3 x^{2} \end{array}$ $\begin{array}{l} (ii) f (x) = (x - 1) (x - 2) \\ = x^{2} - 3 x + 2 \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{{(x + h)}^{2} - 3 (x + h) + 2 - (x^{2} - 3 x + 2)}{h} \\ = \lim_{h \to 0} \frac{x^{2} + 2 xh + h^{2} - 3 x - 3 h + 2 - x^{2} + 3 x - 2}{h} \\ = \lim_{h \to 0} \frac{2 xh + h^{2} - 3 h}{h} \\ = \lim_{h \to 0} \frac{(2 x + h - 3) h}{h} \\ f ‘ (x) = 2 x - 3 \\ (iii) f (x) = \frac{1}{x^{2}} \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{\frac{1}{{(x + h)}^{2}} - \frac{1}{x^{2}}}{h} \\ = \lim_{h \to 0} \frac{x^{2} - {(x + h)}^{2}}{h {(x + h)}^{2} x^{2}} \\ = \lim_{h \to 0} \frac{x^{2} - x^{2} - 2 xh - h^{2}}{h {(x + h)}^{2} x^{2}} \\ = \lim_{h \to 0} \frac{- 2 xh - h^{2}}{h {(x + h)}^{2} x^{2}} \\ = \lim_{h \to 0} \frac{- h (2 x + h)}{h {(x + h)}^{2} x^{2}} \end{array}$ $\begin{array}{l} = \frac{- (2 x + 0)}{{(x + 0)}^{2} x^{2}} \\ = - \frac{2 x}{x^{4}} \\ f ‘ (x) = - \frac{2}{x^{3}} \\ (iv) f (x) = \frac{x + 1}{x - 1} \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{\frac{x + h + 1}{(x + h - 1)} - \frac{x + 1}{(x - 1)}}{h} \\ = \lim_{h \to 0} \frac{(x + h + 1) (x - 1) - (x + 1) (x + h - 1)}{h (x + h - 1) (x - 1)} \\ = \lim_{h \to 0} \frac{(x^{2} + xh + x - x - h - 1) - (x^{2} + xh - x + x + h - 1)}{h (x + h - 1) (x - 1)} \\ = \lim_{h \to 0} \frac{x^{2} + xh + x - x - h - 1 - x^{2} - xh + x - x - h + 1}{h (x + h - 1) (x - 1)} \\ = \lim_{h \to 0} \frac{- 2 h}{h (x + h - 1) (x - 1)} \\ = \frac{- 2}{(x + 0 - 1) (x - 1)} \\ f ‘ (x) = \frac{- 2}{{(x - 1)}^{2}} \end{array}$

Q.15

$For the function f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . .. + \frac{x^{2}}{2} + x + 1 Prove that f ‘ (1) = 100 f ‘ (0) .$

Ans

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

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

Ans

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

Q.17

$Find lim_{x \to 0} f (x),wheref (x) = {\begin{matrix} \frac{x}{| x |}, x \neq 0 \\ 0, x = 0 \end{matrix}$

Ans

$\begin{array}{l} The given function is, \\ f (x) = {\begin{cases} \frac{x}{| x |}, x \neq 0 \\ 0, x = 0 \end{cases} \\ At x = 0, \\ L . H . L . = \lim_{x \to 0^{-}} f (x) \\ = \lim_{x \to 0^{-}} (\frac{x}{| x |}) \\ = \lim_{x \to 0} (\frac{x}{- x}) [when x < 0, | x | = - x] \\ = - 1 \\ R . H . L . = \lim_{x \to 0^{+}} f (x) \\ = \lim_{x \to 0^{+}} (\frac{x}{| x |}) \\ = \lim_{x \to 0} (\frac{x}{x}) [when x < 0, | x | = x] \end{array}$ $\begin{array}{l} = 1 \\ Since, \lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x) \\ So, \lim_{x \to 0} f (x) does not exist. \end{array}$

Q.18

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

Ans

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

Q.19

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

Ans

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

Q.20

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

Ans

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

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

Ans

$\begin{array}{l} Since, f (x) = \cos x \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{h \to 0} \frac{\cos (x + h) - \cos (x)}{h} \\ = \lim_{h \to 0} \frac{- 2 \sin (\frac{x + h - x}{2}) \sin (\frac{x + h + x}{2})}{h} \\ = - \lim_{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \times \lim_{h \to 0} \sin (\frac{2 x + h}{2}) \\ = - 1 \times \sin (\frac{2 x + 0}{2}) \\ f ‘ (x) = - sinx \end{array}$

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

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

Q.23

$\lim_{x \to 0} \frac{e^{4 x} - 1}{x}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{e^{4 x} - 1}{x} = \lim_{4 x \to 0} \frac{e^{4 x} - 1}{4 x} \times 4 (x \to 0 \Rightarrow 4 x \to 0) \\ = 1 \times 4 \\ = 4 \end{array}$

Q.24

$\lim_{x \to 0} \frac{e^{2 + x} - e^{2}}{x}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{e^{2 + x} - e^{2}}{x} = \lim_{x \to 0} \frac{e^{2} (e^{x} - 1)}{x} \\ = e^{2} \lim_{x \to 0} \frac{(e^{x} - 1)}{x} \\ = e^{2} \times 1 [âˆµ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] \\ = e^{2} \end{array}$

Q.25

$\lim_{x \to 0} \frac{e^{x} - e^{5}}{x - 5}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{e^{x} - e^{5}}{x - 5} = \lim_{x \to 0} \frac{e^{5} (e^{x - 5} - 1)}{x - 5} \\ = e^{5} \lim_{x \to 0} \frac{e^{x - 5} - 1}{x - 5} \\ = e^{5} \times 1 [âˆµ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] \\ = e^{5} \end{array}$

Q.26

$\lim_{x \to 0} \frac{e^{sinx} - 1}{x}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{e^{sinx} - 1}{x} = \lim_{sinx \to 0} \frac{e^{sinx} - 1}{sinx} \times \lim_{x \to 0} \frac{sinx}{x} (x \to 0 \Rightarrow sinx \to 0) \\ = 1 \times 1 [\begin{array}{l} âˆµ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1 and \\ \lim_{x \to 0} \frac{sinx}{x} = 1 \end{array}] \\ ∴ \lim_{x \to 0} \frac{e^{sinx} - 1}{x} = 1 \end{array}$

Q.27

$\lim_{x \to 0} \frac{e^{x} - e^{3}}{x - 3}$

Ans

$\lim_{x \to 0} \frac{e^{x} - e^{3}}{x - 3} = \lim_{x \to 0} \frac{e^{3} (e^{x - 3} - 1)}{x - 3}$ $\begin{array}{l} = e^{3} \lim_{x \to 0} \frac{e^{x - 3} - 1}{x - 3} \\ = e^{3} \times 1 [âˆµ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] \\ = e^{3} \end{array}$

Q.28

$\lim_{x \to 0} \frac{x (e^{x} - 1)}{1 - cosx}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{x (e^{x} - 1)}{1 - cosx} = \lim_{x \to 0} \frac{e^{x} - 1}{x} \times \lim_{x \to 0} \frac{x^{2}}{1 - cosx} \\ = 1 \times \lim_{x \to 0} \frac{x^{2}}{2 \sin^{2} (\frac{x}{2})} \\ = \lim_{x \to 0} \frac{2 {(\frac{x}{2})}^{2}}{{\sin (\frac{x}{2})}^{2}} \\ = \lim_{x \to 0} \frac{2}{{\frac{\sin (\frac{x}{2})}{(\frac{x}{2})}}^{2}} \\ = \frac{2}{{(1)}^{2}} \\ = 2 \end{array}$

Q.29

$\lim_{x \to 0} \frac{\log_{e} (1 + 2 x)}{x}$

Ans

$\begin{array}{l} \lim_{x \to 0} \frac{\log_{e} (1 + 2 x)}{x} = \lim_{2 x \to 0} \frac{\log_{e} (1 + 2 x)}{2 x} \times 2 [x \to 0 \Rightarrow 2 x \to 0] \\ = 1 \times 2 [âˆµ \lim_{x \to 0} \frac{\log_{e} (1 + x)}{x} = 1] \\ = 2 \end{array}$

Q.30

$\lim_{x \to 0} \frac{\log (1 + x^{3})}{\sin^{3} x}$

Ans

$\lim_{x \to 0} \frac{\log (1 + x^{3})}{\sin^{3} x} = \lim_{x^{3} \to 0} \frac{\log (1 + x^{3})}{x^{3}} \times \lim_{x^{3} \to 0} \frac{x^{3}}{\sin^{3} x} [âˆµ x \to 0 \Rightarrow x^{3} \to 0]$ $\begin{array}{l} = 1 \times 1 [âˆµ \lim_{x \to 0} \frac{\log (1 + x)}{x} = 1 and \lim_{x \to 0} \frac{x}{sinx} = 1] \\ = 1 \end{array}$

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

$(i) - x (ii) {(- x)}^{- 1} (iii) \sin (x + 1) (iv) \cos (x - \frac{π}{8})$

Ans

$\begin{array}{l} (i) Since, f (x) = - x \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \end{array}$ $\begin{array}{l} = \lim_{h \to 0} \frac{- (x + h) - (- x)}{h} \\ = \lim_{h \to 0} \frac{- x - h + x}{h} \\ = - \lim_{h \to 0} \frac{h}{h} \\ = - 1 \\ (ii) Since, f (x) = {(- x)}^{- 1} = \frac{1}{- x} \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{h \to 0} \frac{- \frac{1}{(x + h)} - (\frac{1}{- x})}{h} \\ = \lim_{h \to 0} \frac{- \frac{1}{(x + h)} + \frac{1}{x}}{h} \\ = \lim_{h \to 0} \frac{\frac{- x + x + h}{x (x + h)}}{h} \\ = \lim_{h \to 0} \frac{h}{hx (x + h)} \\ = \frac{1}{x (x + 0)} \\ = x^{- 2} \\ (iii) Since, f (x) = \sin (x + 1) \end{array}$ $\begin{array}{l} ∴ f ‘ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{h \to 0} \frac{\sin (x + h + 1) - \sin (x + 1)}{h} \\ = \lim_{h \to 0} \frac{2 \cos (\frac{x + h + 1 + x + 1}{2}) \sin (\frac{x + h + 1 - x - 1}{2})}{h} \\ = \lim_{h \to 0} \frac{2 \cos (\frac{2 x + h + 2}{2}) \sin (\frac{h}{2})}{h} \\ = \lim_{h \to 0} \cos (\frac{2 x + h + 2}{2}) \times \lim_{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} [h \to 0 \Rightarrow \frac{h}{2} \to 0] \\ = \cos (\frac{2 x + 0 + 2}{2}) \times 1 \\ = \cos (x + 1) \\ (iv) Since, f (x) = \cos (x - \frac{π}{8}) \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{h \to 0} \frac{\cos (x + h - \frac{π}{8}) - \cos (x - \frac{π}{8})}{h} \\ = \lim_{h \to 0} \frac{- 2 \sin (\frac{x + h - \frac{π}{8} + x - \frac{π}{8}}{2}) \sin (\frac{x + h - \frac{π}{8} - x + \frac{π}{8}}{2})}{h} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = - \lim_{h \to 0} \sin (\frac{2 x + h - \frac{2 π}{8}}{2}) \times \lim_{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} [h \to 0 \Rightarrow \frac{h}{2} \to 0] \\ = - \sin (\frac{2 x + 0 - \frac{2 π}{8}}{2}) \times 1 \\ = - \sin (x - \frac{π}{8}) \end{array}$

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

Ans

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

Q.33

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

Ans

$\begin{array}{l} Let f (x) = (px + q) (\frac{r}{x} + s) \\ Differentiating w.r.t. x, we get \end{array}$ $\begin{array}{l} f’ (x) = \frac{d}{dx} {(px + q) (\frac{r}{x} + s)} \\ = (px + q) {\frac{d}{dx} (\frac{r}{x} + s)} + (\frac{r}{x} + s) {\frac{d}{dx} (px + q)} \\ = (px + q) (r \frac{d}{dx} x^{- 1} + \frac{d}{dx} s) + (\frac{r}{x} + s) (p \frac{d}{dx} x + \frac{d}{dx} q) \\ = (px + q) (- {rx}^{- 2} + 0) + (\frac{r}{x} + s) (p \times 1 + 0) \\ = - \frac{pr}{x} - {qrx}^{- 2} + \frac{pr}{x} + ps \\ = - {qrx}^{- 2} + ps \\ = - \frac{qr}{x^{2}} + ps \\ ∴ f ‘ (x) = - \frac{qr}{x^{2}} + ps \end{array}$

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

Ans

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

Q.35

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

Ans

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

Q.36

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

Ans

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

Q.37

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

Ans

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

Q.38

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

Ans

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

Q.39

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

Ans

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

Q.40

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

Ans

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

Q.41

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

Ans

$\begin{array}{l} Let f (x) = 4 \sqrt{x} - 2 \\ = 4 x^{\frac{1}{2}} - 2 \end{array}$ $\begin{array}{l} Differentiating w.r.t. x, we get \\ f’ (x) = \frac{d}{dx} (4 x^{\frac{1}{2}} - 2) \\ = 4 \frac{d}{dx} x^{\frac{1}{2}} - \frac{d}{dx} (2) \\ = 4 \times \frac{1}{2} x^{- \frac{1}{2}} - 0 \\ = 2 x^{- \frac{1}{2}} \\ = \frac{2}{\sqrt{x}} \end{array}$

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

Ans

$\begin{array}{l} Let f (x) = {(ax + b)}^{n} and f (x + h) = {\{a (x + h) + b\}}^{n} \\ = {(ax + ah + b)}^{n} \\ By first Principle, f’ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ \frac{d}{dx} {(ax + b)}^{n} = \lim_{h \to 0} \frac{{(ax + ah + b)}^{n} - {(ax + b)}^{n}}{h} \\ = \lim_{h \to 0} \frac{{(ax + b)}^{n} \{{(\frac{ax + ah + b}{ax + b})}^{n} - 1\}}{h} \\ = \lim_{h \to 0} \frac{{(ax + b)}^{n} \{{(\frac{ax + b}{ax + b} + \frac{ah}{ax + b})}^{n} - 1\}}{h} \\ = \lim_{h \to 0} \frac{{(ax + b)}^{n} \{{(1 + \frac{ah}{ax + b})}^{n} - 1\}}{h} \\ = \lim_{h \to 0} \frac{{(ax + b)}^{n} \{1 + \frac{nah}{ax + b} + \frac{n (n - 1) a^{2} h^{2}}{2! (ax + b)} + . .. - 1\}}{h} \\ [Using binomial theorem] \\ = \lim_{h \to 0} {(ax + b)}^{n} \{\frac{na}{ax + b} + \frac{n (n - 1) a^{2} h}{2! (ax + b)} + . ..\} \\ = {(ax + b)}^{n} \{\frac{na}{ax + b} + 0 + . ..\} \\ = \frac{na}{ax + b} \times {(ax + b)}^{n} \\ \frac{d}{dx} {(ax + b)}^{n} = na {(ax + b)}^{n - 1} \end{array}$

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

Ans

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

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

Ans

$\begin{array}{l} Let f (x) = \sin (x + a) and f (x + h) = \sin (x + h + a) \\ By first Principle, \\ f’ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{h \to 0} \frac{\sin (x + h + a) - \sin (x + a)}{h} \\ = \lim_{h \to 0} \frac{2 \cos (\frac{x + h + a + x + a}{2}) \sin (\frac{x + h + a - x - a}{2})}{h} \\ = \lim_{h \to 0} \cos (\frac{2 x + h + 2 a}{2}) \times \lim_{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} [h \to 0 \Rightarrow \frac{h}{2} \to 0] \end{array}$ $\begin{array}{l} = \cos (\frac{2 x + 0 + 2 a}{2}) \times 1 \\ = \cos (x + a) \times 1 \\ = \cos (x + a) \\ Therefore, \\ \frac{d}{dx} \sin (x + a) = \cos (x + a) \end{array}$

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

Ans

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

Q.46

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

Ans

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

Q.47

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

Ans

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

Q.48

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

Ans

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

Q.49 Find the derivative of the function sinⁿx.

Ans

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

Q.50

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

Ans

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

Q.51

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

Ans

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

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

Ans

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

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

Ans

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

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

Ans

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

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

Ans

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

Q.56

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

Ans

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

Q.57

$Find the derivative of the function \frac{x^{2} cos (\frac{π}{4})}{sinx} .$

Ans

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

Q.58

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

Ans

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

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

Ans

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

Q.60

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

Ans

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

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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.

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

They must read each question carefully and attentively. Analyse the type of problem.
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They must solve it in a step-by-step format.
They must begin with easy steps and calculations.
They must double-check from the start once the calculation is done.

NCERT Solutions Class 11 Maths Chapter 13

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

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

NCERT Solutions for Class 11 Mathematics Chapter 13 Exercise & Solutions

NCERT Exemplar Class 11 Mathematics

Key Features for NCERT Solutions for Class 11 Mathematics Chapter 13

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NCERT Solutions Class 11 Maths Chapter 13

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

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

NCERT Solutions for Class 11 Mathematics Chapter 13 Exercise & Solutions

NCERT Exemplar Class 11 Mathematics

Key Features for NCERT Solutions for Class 11 Mathematics Chapter 13

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