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NCERT Solutions Class 12 Mathematics Chapter 5 – Continuity and Differentiability

NCERT Solutions Class 12 Mathematics Chapter 5 Continuity and Differentiability include all concepts and theories related to the said topic. Students will learn and understand how to implement these concepts. In addition, they will also learn the rigorous formulation of the intuitive concept of function that varies. With help from NCERT Solutions Class 12 Mathematics Chapter 5, students can understand the complex theories and the simplest way to solve the problems.

NCERT Solutions Class 12 Mathematics Chapter 5 covers essential topics such as exponential and logarithmic functions. The subtopics include continuity, differentiability, logarithmic differentiation, and mean value theorem. In addition, they will engage on the core topic of the algebra of continuous functions. The chapter carries essential formulae and helps to improve their calculation. The study material is available online on Extramarks. Extramarks study material and revision notes offer solutions for all problems and answers to the questions enlisted under the chapter.

Under Continuity and Differentiability, students will easily understand the complex theorems with help from Extramarks NCERT Solutions Class 12 Mathematics Chapter 5. Furthermore, as the solutions are built based on CBSE NCERT’s latest 2022-2023 syllabus, students can be assured that they are up-to-date and ready to tackle anything in their examination. Students can visit the Extramarks website for the latest news and updates regarding Class 12 Mathematics Chapter 5.

Key Topics Covered In NCERT Solutions for Class 12 Mathematics Chapter 5

NCERT Solutions for Class 12 Mathematics Chapter 5 Continuity and Differentiability elaborates the fundamental concepts. It includes continuity of functions, their differentiability, and their relations. In addition, the notion of the continuity of a function is applied in various concepts and theories. Thus, students may refer to NCERT Solutions Class 12 Mathematics Chapter 5 to be more thorough on various concepts under the chapter. With daily practice, students can quickly gain a deep understanding of the topic. Some of the essential topics elaborated in the chapter are mainly seen repeating in various entrance exams. It includes logarithmic differentiation, derivatives of functions, and second-order derivatives.

Key topics covered in Continuity and Differentiation include

Exercise	Topics
5.1	Introduction
5.2	Continuity
5.3	Differentiability
5.4	Exponential and Logarithmic Functions
5.5	Logarithmic Differentiation
5.6	Derivatives of Functions in Parametric Forms
5.7	Second-Order Derivative
5.8	Mean Value Theorem

5.1 Introduction

The NCERT Solutions Class 12 Mathematics Chapter 5 covers all important topics and offers step-by-step solutions. Students will get an introduction to Continuity and Differentiability. It has advanced theorems and logarithms. They will also learn the essential concepts of continuity, differentiability, and relations. Further, they can grasp the knowledge of inverse trigonometric functions.

5.2 Continuity

This section is about continuity. There are various formulaeand concepts to study and understand. The students will engage in learning algebra of continuous functions. They will also learn analogous algebraic uninterrupted functions. The continuity of a function at a point is expected to show similar results to the case of limits. With the help of NCERT Solutions Class 12 Mathematics Chapter 5, the students will learn basic definitions and theorems.

5.3 Differentiability

Differentiability includes a list of derivatives and some important theorems. It is essential to know that there are a total of three theorems that appear in competitive exams. The students will learn about the derivatives of composite, implicit, and inverse trigonometric functions. They will also get in-depth learning about derivatives of implicit function.

5.4 Exponential and Logarithmic Functions

In NCERT Solutions Class 12 Mathematics, Chapter 5 includes all important sub-topics under Exponential and Logarithmic Functions. The students will witness different classes of functions like polynomial functions, rational functions, and trigonometric functions. They will also learn about a new class of related and logarithmic functions. This exercise emphasises the exponential functions and logarithms. Several examples provided for their reference help the students grasp and understand exponential and logarithmic functions.

5.5 Logarithmic Differentiation

Students learn all essential theories in NCERT Solutions Class 12 Mathematics Chapter 5 Continuity and Differentiability. The students will learn to differentiate the specific class of functions given in the particular forms. Logarithmic Differentiation is an essential topic as it appears in entrance exams such as JEE Mains and JEE Advanced.

5.6 Derivatives of Functions in Parametric Forms

The students will learn about different concepts, the relation between two variables, and the third variables in this exercise. The NCERT Solutions Class 12 Mathematics Chapter 5 covers every problem and offers a step-by-step solution. The students will be given several examples to understand the topic better and grasp the concepts.

5.7 Second Order Derivative

The students will learn how to work with second-order derivatives and their applications in various theorems. There are more examples and solutions than theory, as this topic requires a more practical approach than a theoretical approach. These examples will help the students to get a hold of the theorems and concepts.

5.8 Mean Value Theorem

In Mean Value Theorem, students will learn the functionality of the mean value theorem and its relations to Rolle’s theorem. Various examples help the students to engage with the theorem, and it’s a core concept. The mean value theorem is slightly tricky. However, students can refer to NCERT Solutions Class 12 Mathematics Chapter 5. It provides a step-by-step solution to an example based on the mean value theorem.

In addition to the details mentioned above, the chapter also includes formulaeand a detailed explanation of the formulae. Some of the formulaeof various functions in NCERT Solutions Chapter 5 Mathematics Class 12 include

Arithmetic of Continuous Functions: The arithmetic operations performed on continuous functions are said to be continuous. If f and g both are continuous functions, then:

(f ± g) (x) = f(x) ± g(x) is continuous.

(f . g) (x) = f(x) . g (x) is continuous.

( f / g ) (x) = f(x ) / g(x) (wherever g(x) ≠ 0) is continuous.

Logarithmic Functions: The logarithmic functions are also known as exponential functions. If the logarithmic function y = ax is equivalent, it is equivalent to the exponential equation x = ay.

Students may click here to access the NCERT Solutions for Class 12 Mathematics Chapter 5 on Extramarks.

NCERT Solutions Class 12 Mathematics Chapter 5 Exercise & Answer Solutions

NCERT Solutions Class 12 Mathematics Chapter 5 Exercise and Answer Solutions are available for students to refer for free on the Extramarks website. The material offers step-by-step solutions that help students understand how to solve problems relating to the chapter. The solution also helps students solve complex questions in a simplified manner. Students may refer to NCERT Solutions Class 12 Mathematics Chapter 5 to strengthen their basics.

Students may refer to the links below to download exercise-specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 5 Continuity and Differentiation:

Chapter 5: Exercise 5.1 Solutions – 34 Questions (10 Short Questions and 24 Long Questions)
Chapter 5: Exercise 5.2 Solutions – 10 Questions (2 Short Questions 8 Long Questions)
Chapter 5: Exercise 5.3 Solutions – 15 Questions (9 short Questions and 6 long Questions)
Chapter 5: Exercise 5.4 Solutions – 10 Questions (5 Short Questions and 5 Long Questions)
Chapter 5: Exercise 5.5 Solutions – 18 Questions (4 Short Questions and 14 Long Questions)
Chapter 5: Exercise 5.6 Solutions – 11 Questions (1 Short Question and 1 Long Questions)
Chapter 5: Exercise 5.7 Solutions – 17 Questions (10 Short Questions and 7 Long Questions)
Chapter 5: Exercise 5.8 Solutions – 6 Questions (3 Short Questions and 3 Long Questions)

Students can also download other NCERT Solutions for all chapters of class 12 Mathematics on the Extramarks website. In addition, students can also find study material for different classes, as mentioned below.

NCERT Solutions Class 1
NCERT Solutions Class 2
NCERT Solutions Class 3
NCERT Solutions Class 4
NCERT Solutions Class 5
NCERT Solutions Class 6
NCERT Solutions Class 7
NCERT Solutions Class 8
NCERT Solutions Class 9
NCERT Solutions Class 10
NCERT Solutions Class 11.

Students can click on the links to refer to the study material. In addition to the study material, Extramarks also provides sample papers and a list of important questions based on past year question papers of at least a decade. All information can be accessed for free by clicking on the link mentioned above.

NCERT Exemplar Class 12 Mathematics

The NCERT Exemplar Class Mathematics is an excellent source of information. It helps the students to get a grasp on the essential and complex concepts. Furthermore, the topics present in NCERT are important as they appear in the entrance exams. Therefore, students can refer to the exemplar to score better marks. It can be referred to especially when the exams are neat, and the student is short of time. Thus, an exemplar can save time and provide the best subject knowledge.

The Chapter 5 Class 12 Mathematics exemplar books have more information and slightly complex questions. The exemplars help the students practise more and prepare them for competitive exams such as JEE Main and JEE Advanced. Extramarks understand the importance of such CBSE and NCERT resources. Thus, students can download the exemplar from the Extramarks website.

Key Features of NCERT Solutions Class 12 Mathematics Chapter 5

NCERT Solutions Class 12 Mathematics Chapter 5 offers proper solutions for every example problem. With the help of NCERT Solutions, the students can score more and get a proper understanding of complex theories. Some of the key features of NCERT Solutions Class 12 Mathematics Chapter 5 include

Every topic in Chapter 5 Class 12 Mathematics offers an in-depth explanation that helps the students understand the concept.
The solutions are based on the latest syllabus of the CBSE 2022-23 and aid in preparing for the examination.
Students can expand their knowledge in the core topics and get more practice solving problems.
The students can understand their assignments based on concepts from continuity and differentiability.
Class 12 Mathematics Chapter 5 has some important elements that can help prepare for various competitive exams.

Students can benefit from referring to the NCERT Solutions Class 12 Mathematics Chapter 5 by clicking here.

Q.1

$\begin{array}{l} Prove that the function f (x) = 5 x - 3 is continuous at x = 0, \\ at x = - 3 and at x = 5 . \end{array}$

Ans.

$\begin{array}{l} Given function is f (x) = 5 x - 3 \\ At x = 0, f (0) = 5 (0) - 3 = - 3 \\ \lim_{x \to 0} f (x) = \lim_{x \to 0} 5 x - 3 \\ = 5 (0) - 3 \\ = - 3 \\ ∴ \lim_{x \to 0} f (x) = f (0) \\ Therefore, f is continuous at x = 0 . \\ At x = - 3, f (- 3) = 5 (- 3) - 3 = - 18 \\ \lim_{x \to - 3} f (x) = \lim_{x \to 0} 5 x - 3 \\ = 5 (- 3) - 3 \\ = - 18 \\ ∴ \lim_{x \to - 3} f (x) = f (- 3) \\ Therefore, f is continuous at x = - 3 . \\ At x = 5, f (5) = 5 (5) - 3 = 22 \\ \lim_{x \to 5} f (x) = \lim_{x \to 0} 5 x - 3 \\ = 5 (5) - 3 \\ = 22 \\ ∴ \lim_{x \to 5} f (x) = f (5) \\ Therefore, f is continuous at x = 5 . \end{array}$

Q.2 Examine the continuity of the function
f(x) = 2x² – 1 at x = 3.

Ans.

$\begin{array}{l} The given function is f (x) = 2 x^{2} - 1 \\ At x = 3, f (3) = 2 {(3)}^{2} - 1 = 17 \\ \lim_{x \to 3} f (x) = \lim_{x \to 3} 2 x^{2} - 1 \\ = 2 {(3)}^{2} - 1 \\ = 17 \\ ∴ \lim_{x \to 3} f (x) = f (3) \\ Therefore, f is continuous at x = 3 . \end{array}$

Q.3

$\begin{array}{l} Examine the following functions for continuity . \\ (a) f (x) = x - 5 (b) f (x) = \frac{1}{x - 5}, x \neq 3 \\ (c) f (x) = \frac{x^{2} - 25}{x + 5}, x \neq - 5 (d) f (x) = |x - 5| \end{array}$

Ans.

$\begin{array}{l} (a) The given function is f (x) = x - 5 \\ It is evident that f is defined at every real number k, \\ then f (k) = k - 5 . \\ \lim_{x \to k} f (x) = \lim_{x \to k} (x - 5) = k - 5 \\ ∴ \lim_{x \to k} f (x) = f (k) \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \\ (b) The given function is f (x) = \frac{1}{x - 5}, x \neq 5 \\ For any real number k \neq 5 \\ then f (k) = k - 5 . \\ \lim_{x \to k} f (x) = \lim_{x \to k} (x - 5) = k - 5 \\ ∴ \lim_{x \to k} f (x) = f (k) \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \\ (c) The given function is f (x) = \frac{x^{2} - 25}{x + 5}, x \neq - 5 \\ For any real number k \neq - 5 \\ then f (k) = \frac{(k - 5) (k + 5)}{(k + 5)} = k - 5 \\ \lim_{x \to - k} f (x) = \lim_{x \to k} \frac{(k - 5) (k + 5)}{(k + 5)}) = k - 5 \\ ∴ \lim_{x \to k} f (x) = f (k) \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \\ (d) The given function is f (x) = | x - 5 | = {\begin{cases} - x + 5, if x < 5 \\ x - 5, if x \geq 5 \end{cases} \\ This function f is defined at all points of the real line . \\ Let c be a point on a real line . Then, k < 5 or k = 5 or k > 5 \\ Case I : c < 5 \\ Then, f (c) = 5 - c \\ \lim_{x \to c} f (x) = \lim_{x \to c} (5 - x) \\ = 5 - c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all real numbers less than 5 . \\ Case II : c = 5 \\ Then, f (c) = f (5) \\ = 5 - 5 = 0 \\ \lim_{x \to 5^{-}} f (x) = \lim_{x \to 5} (5 - x) \\ = 5 - 5 = 0 \\ \lim_{x \to 5^{+}} f (x) = \lim_{x \to 5} (x - 5) \\ = 5 - 5 = 0 \\ ∴ \lim_{x \to 5^{-}} f (x) = \lim_{x \to 5^{+}} f (x) = f (c) \\ Therefore, f is continuous at x = 5 \\ Case III : c > 5 \\ Then, f (c) = c - 5 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x - 5) = c - 5 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all real numbers greater than 5 . \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \end{array}$

Q.4

$\begin{array}{l} P r o v e t h a t t h e f u n c t i o n f (x) = x^{n} i s c o n t i n u o u s a t x = n, \\ w h e r e n i s a p o s i t i v e i n t e g e r . \end{array}$

Ans.

$\begin{array}{l} The given function is f (x) = x^{n} \\ It is evident that f is defined at all positive integers, n, and \\ its value at n is n^{n} . \\ Then, \lim_{x \to n} f (n) = \lim_{x \to n} x^{n} = n^{n} \\ ∴ \lim_{x \to n} f (x) = f (n) \\ Therefore, f is continuous at n, where n is a positive integer . \end{array}$

Q.5

$\begin{array}{l} Is the function f defined by \\ f (x) = \{\begin{cases} x, if x \leq 1 \\ 5, if x > 1 \end{cases} \\ continuous at x = 0 ? Atx = 1 ? Atx = 2 ? \end{array}$

Ans.

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} x, if x \leq 1 \\ 5, if x >1 \end{cases} \\ At x =0 \\ It is evident that f is defined at 0 and its value at 0 is 0 . \\ Then, \lim_{x \to 0} f (x) = \lim_{x \to 0} x = 0 \\ ∴ \lim_{x \to 0} f (x) = f (0) \\ Therefore, f is continuous at x = 0 \\ At x = 1, \\ f is defined at 1 and its value at 1 is 1 . \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} x = 1 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (5) = 5 \\ ∴ \lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x) \\ Therefore, f is not continuous at x = 1 . \\ At x = 2, \\ f is defined at 2 and its value at 2 is 5 . \\ Then, \lim_{x \to 2} f (x) = \lim_{x \to 2} (5) = 5 \\ ∴ \lim_{x \to 2} f (x) = f (2) \\ Therefore, f is continuous at x = 2 . \end{array}$

Q.6 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} 2 x + 3, if x \leq 2 \\ 2 x - 3, if x > 2 \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} 2x+3, if x \leq 2 \\ 2x - 3, if x>2 \end{cases} \\ It is evident that the given function f is defined at all \\ the points of the real line . \\ Let c be a point on the real line . Then, three cases arise . \\ (i) c < 2 \\ (i i) c > 2 \\ (i i i) c = 2 \\ C a s e 1 : c < 2 \\ T h e n, f (c) = 2 c + 3 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2x+3) = 2c+3 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 2 . \\ C a s e 2 : c > 2 \\ T h e n, f (c) = 2 c - 3 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2x - 3) = 2c - 3 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 2 . \\ Case (iii) c = 2 \\ Then, the left hand limit of f at x = 2 is, \\ \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2} (2x+3) = 7 \\ \lim_{x \to 2^{+}} f (x) = \lim_{x \to 2} (2x - 3) = 1 \\ ∴ \lim_{x \to 2^{-}} f (x) \neq \lim_{x \to 2^{+}} f (x) \\ Therefore, f is not continuous at x = 2 \\ Hence, x = 2 is the only point of discontinuity of f . \end{array}$

Q.7 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} | x | + 3, if x \leq - 3 \\ - 2 x, if - 3 < x < 3 \\ 6 x + 2, if x \geq 3 \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} | x | + 3 = - x +3, if x \leq - 3 \\ - 2 x, if - 3 < x < 3 \\ 6 x + 3, if x \geq 3 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case 1 : If c < - 3 \\ Then, f (c) = - c + 3 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (- x +3) = - c +3 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < - 3 . \\ Case 2 : c = - 3 \\ Then, f (- 3) = - (- 3) + 3 = 6 \\ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{-}} (- x +3) = - (- 3) + 3 = 6 \\ \lim_{x \to 3^{+}} f (x) = \lim_{x \to 3^{+}} (- 2 x) = - 2 (- 3) = 6 \\ ∴ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{+}} f (x) = f (- 3) \\ Therefore, f is continuous at x = - 3. \\ Case 3 : If - 3 < c < 3, then \\ f (c) = - 2 c and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (- 2 x) = - 2 c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous in (- 3,3) . \\ Case 4: If c = 3, then \\ L . H . L . = \lim_{x \to 3^{-}} f (x) \\ = \lim_{x \to 3^{-}} (- 2 x) \\ = - 2 (- 3) = 6 \\ R . H . L . = \lim_{x \to 3^{+}} f (x) \\ = \lim_{x \to 3^{+}} (6 x +2) \\ = 6 (3) + 2 = 20 \\ ∴ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{+}} f (x) \\ Therefore, f is not continuous at x = 3 \\ Case 5: If c > 3, then \\ f (c) = 6 c + 2 and \lim_{x \to c} f (x) = \lim_{x \to c} (6 x + 2) = 6 c + 2 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 3 \\ Hence, x = 3 is the only point of discontinuity of f . \end{array}$

Q.8 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} \frac{| x |}{x}, if x \neq 0 \\ 0, if x = 0 \end{array}$

Ans.

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} \frac{| x |}{x}, if x \neq 0 \\ 0, if x =0 \end{cases} \\ = {\begin{cases} \frac{| x |}{x} = \frac{x}{x} = 1, if x > 0 \\ 0, if x = 0 \\ \frac{| x |}{x} = \frac{- x}{x} = - 1, if x < 0 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <0, then f (c) = - 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (- 1) = - 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x < 0 . \\ Case II : If c = 0, then L . H . L . at x = 0 is, \\ L . H . L . = \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (- 1) = - 1 \\ L . H . L . at x = 0 ‹ is \\ R . H . L . = \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (1) = 1 \\ L . H . L . \neq R . H . L . \\ Therefore, f is not continuous at x = 0 \\ Case III : If c > 0, then f (c) = 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (1) = 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x > 0 . \\ Hence, x = 0 is the only point of discontinuity of f . \end{array}$

Q.9 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} \frac{x}{| x |}, if x < 0 \\ - 1, if x \geq 0 \end{array}$

Ans.

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} \frac{x}{| x |} = \frac{x}{- x} = - 1, if x < 0 \\ - 1, if x \geq 0 \end{cases} \\ \Rightarrow f (x) = - 1 for all x \in R \\ Let c be any real number . Then, \lim_{x \to c} f (x) = \lim_{x \to c} (- 1) = - 1 \\ And f (c) = - 1 for all x \in R \\ Therefore, the given function is a continuous function . \\ Hence, the given function has no point of discontinuity . \end{array}$

Q.10 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} x + 1, if x \geq 1 \\ x^{2} + 1, if x < 1 \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x +1, if x \geq 1 \\ x^{2} + 1, if x <1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c^{2} + 1 ‹‹ ‹ and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + 1) \\ = c^{2} + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, then f (c) = f (1) = 1 + 1 = 2 \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} + 1) = 1^{2} + 1 = 2 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x + 1) = 1 + 1 = 2 \\ ∴ \lim_{x \to 1} f (x) = f (1) \\ Therefore, f is continuous at x = 1 . \\ Case III : If c > 1, then f (c) = c + 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 1) \\ = c + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Hence, the given function f has no point of discontinuity . \end{array}$

Q.11 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} x^{3} - 3, if x \leq 2 \\ x^{2} + 1, if x > 2 \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x^{3} - 3, if x \leq 2 \\ x^{2} + 1, if x >2 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c^{2} + 1 ‹‹ ‹ and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + 1) \\ = c^{2} + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, then f (c) = f (1) = 1 + 1 = 2 \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} + 1) = 1^{2} + 1 = 2 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x + 1) = 1 + 1 = 2 \\ ∴ \lim_{x \to 1} f (x) = f (1) \\ Therefore, f is continuous at x = 1 . \\ Case III : If c > 1, then f (c) = c + 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 1) \\ = c + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Hence, the given function f has no point of discontinuity . \\ The given function f is f (x) = {\begin{cases} x +1, if x \geq 1 \\ x^{2} + 1, if x <1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c^{2} + 1 ‹‹ ‹ and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} + 1) \\ = c^{2} + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, then f (c) = f (1) = 1 + 1 = 2 \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} + 1) = 1^{2} + 1 = 2 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x + 1) = 1 + 1 = 2 \\ ∴ \lim_{x \to 1} f (x) = f (1) \\ Therefore, f is continuous at x = 1 . \\ Case III : If c > 1, then f (c) = c + 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 1) \\ = c + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Hence, the given function f has no point of discontinuity . \end{array}$

Q.12 Find all the points of discontinuity of f, where f is defined by

$f(x)={\begin{array}{l} x^{10} - 1, if x \leq 1 \\ x^{2}, if x > 1 \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x^{10} - 1, if x \geq 1 \\ x^{2}, if x <1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c^{10} - 1 ‹‹ ‹ and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{10} - 1) \\ = c^{10} - 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{10} - 1) = 1^{10} - 1 = 0 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x^{2}) = 1^{2} = 1 \\ ∴ \lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x) \\ It is observed that the left and right hand limit of f at x = 1 do \\ not coincide . Therefore, f is not continuous at x = 1 \\ Case III : If c > 1, then f (c) = c^{2} \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2}) \\ = c^{2} \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Thus, from the above observation, it can be concluded that \\ x = 1 is the only point of discontinuity of f . \end{array}$

Q.13

$\begin{array}{l} Is the function defined by \\ f (x)= \{\begin{matrix} x + 5 & if x \leq 1 \\ x - 5 & if x >1 \end{matrix} \\ a continuous function ? \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x + 5, if x \leq 1 \\ x - 5, if x >1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <1, then f (c) = c + 5 ‹‹ ‹ and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 5) \\ = c + 5 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 . \\ Case II : If c = 1, then f (1) = 1 + 5 = 6 \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x + 5) = 1 + 5 = 6 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x - 5) = 1 - 5 = - 4 \\ ∴ \lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x) \\ Therefore, f is not continuous at x = 1 \\ Case III : If c > 1, then f (c) = c - 5 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x - 5) \\ = c - 5 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Thus, from the above observation, it can be concluded that \\ x = 1 is the only point of discontinuity of f . \end{array}$

Q.14 Discuss the continuity of the function f, where f is defined by

$f(x)={\begin{array}{l} 3, if 0 \leq x \leq 1 \\ 4, if 1 < x < 3 \\ 5, if 3 \leq x \leq 10 \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} 3, & i f 0 \leq x \leq 1 \\ 4, & i f 1 < x < 3 \\ 5, & i f 3 \leq x \leq 10 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ C a s e I : I f 0 \leq c < 1, t h e n f (c) = 3 a n d \\ lim_{x \to c} â¡ f (x) = lim_{x \to c} (3) \\ = 3 \\ ∴ lim_{x \to c} â¡ f (x) = f (c) \\ Therefore, f is continuous in the interval [0,1) . \\ C a s e I I : I f c = 1, t h e n f (1) = 3 \\ The left hand limit of f at x = 1 is, \\ lim_{x \to 1^{-}} â¡ f (x) = lim_{x \to 1^{-}} (3) = 3 \\ The right hand limit of f at x = 1 is, \\ lim_{x \to 1^{+}} â¡ f (x) = lim_{x \to 1^{+}} (4) = 4 \\ ∴ lim_{x \to 1^{-}} â¡ f (x) \neq lim_{x \to 1^{+}} â¡ f (x) \\ Therefore, f is not continuous at x = 1 \\ C a s e I I I : I f 1 < c < 3, t h e n f (c) = 4 a n d \\ lim_{x \to c} â¡ f (x) = lim_{x \to c} (4) \\ = 4 \\ ∴ lim_{x \to c} â¡ f (x) = f (c) \\ Therefore, f is continuous at all points of the interval (1, 3) . \\ C a s e I V : I f c = 3, t h e n f (c) = 5 \\ T h e l e f t h a n d l i m i t o f f a t x = 3 i s, \\ lim_{x \to 3^{-}} â¡ f (x) = lim_{x \to 3^{-}} (4) = 4 \\ The right hand limit of f at x = 3 is, \\ lim_{x \to 3^{+}} â¡ f (x) = lim_{x \to 3^{+}} (5) = 5 \\ ∴ lim_{x \to 3^{-}} â¡ f (x) \neq lim_{x \to 3^{+}} â¡ f (x) \\ Therefore, f is not continuous at x = 3 . \\ C a s e I V : I f 3 < c \leq 10, t h e n f (c) = 5 a n d \\ lim_{x \to c} â¡ f (c) = lim_{x \to c} (5) = 5 \\ lim_{x \to c} â¡ f (c) = f (c) \\ Therefore, f i s c o n t i n u o u s a t a l l p o i n t s o f t h e i n t e r v a l (3, 10] . \\ H e n c e, f i s n o t c o n t i n u o u s a t x = 1 a n d x = 3 . \end{array}$

Q.15 Discuss the continuity of the function f, where f is defined by

$f(x)={\begin{array}{l} 2 x, if x < 0 \\ 0, if 0 \leq x < 1 \\ 4 x, if x > 1 \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} 2 x, if x < 0 \\ 0, if 0 \leq x <1 \\ 4 x, if x > 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c >0, then f (c) = 2 c ‹‹ \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2 x) \\ = 2 c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 0 . \\ Case II : If c = 0, then f (c) = f (0) = 0 \\ The left hand limit of f at x = 0 is, \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (2 x) = 2 \times 0 = 0 \\ The right hand limit of f at x = 0 is, \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (0) = 0 \\ ∴ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (c) \\ Therefore, f is continuous at x = 0 . \\ Case III : If 0< c < 1, then f (c) = 0 . \\ \lim_{x \to c} f (x) = \lim_{x \to c} (0) = 0 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points of the interval (0, 1) . \\ Case IV : If c = 1, then f (c) = f (1) = 0 . \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (0) = 0 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (4 x) = 4 \times 1 = 4 \\ \Rightarrow \lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x) \\ Therefore, f is not continuous at x = 1 . \\ Case V : If c < 1, then f (c) = 4 c and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (4 x) = 4 c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 1 \\ Hence, f is not continuous only at x = 1 . \end{array}$

Q.16 Discuss the continuity of the function f, where f is defined by

$f(x)={\begin{array}{l} - 2, if x < - x \\ 2 x, if 0 \leq x < 1 \\ 2, if x > 1 \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} - 2, if x \leq - 1 \\ 2 x, if - 1 \leq x <1 \\ 2, if x > 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c < - 1, then f (c) = - 2 and ‹‹ \\ \lim_{x \to c} f (x) = \lim_{x \to c} (- 2) \\ = - 2 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < - 1 . \\ Case II : If c = - 1, then f (c) = f (- 1) = - 2 \\ The left hand limit of f at x = - 1 is, \\ \lim_{x \to - 1^{-}} f (x) = \lim_{x \to - 1^{-}} (- 2) = - 2 \\ The right hand limit of f at x = - 1 is, \\ \lim_{x \to - 1^{+}} f (x) = \lim_{x \to - 1^{+}} (2 x) = 2 \times (- 1) = - 2 \\ ∴ \lim_{x \to - 1} f (x) = f (- 1) \\ Therefore, f is continuous at x = - 1. \\ Case III : If - 1 < c < 1, then f (c) = 2 c . \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2 x) = 2 c \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points of the interval (-1, 1) . \\ Case IV : If c = 1, then f (c) = f (1) = 2 \times 1 = 2 . \\ The left hand limit of f at x = 1 is, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (2 x) = 2 \times 1 = 2 \\ The right hand limit of f at x = 1 is, \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (2) = 2 \\ \Rightarrow \lim_{x \to 1} f (x) = f (c) \\ Therefore, f is not continuous at x = 1 . \\ Case V : If c > 1, then f (c) = 2 and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (2) = 2 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 1 \\ Thus, from the above observations, it can be concluded that \\ f is continuous at all points of the real line . \end{array}$

Q.17

$\begin{array}{l} Find the relationship between a and b so that the function \\ f defined by \\ f (x) = \{\begin{cases} ax + 1, if x \leq 3 \\ bx +, if x > 3 \end{cases} \\ tinuous at x = 3 . \end{array}$

Ans.

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} ax + 1, if x \leq 3 \\ bx + 3, if x > 3 \end{cases} \\ If f is continuous at x = 3, then \\ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{+}} f (x) = f (3) . .. (i) \\ \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{-}} (ax + 1) = 3 a + 1 \\ \lim_{x \to 3^{+}} f (x) = \lim_{x \to 3^{+}} (bx + 3) = 3 b + 3 \\ f (3) = 3 a + 1 \\ Therefore, from (1), we obtain \\ 3 a + 1 = 3 b + 3 = 3 a + 1 \\ \Rightarrow 3 a + 1 = 3 b + 3 \\ \Rightarrow 3 a = 3 b + 3 - 1 \\ \Rightarrow a = b + \frac{2}{3} \\ Therefore, the required relationship is given by \\ a = b + \frac{2}{3} . \end{array}$

Q.18

$\begin{array}{l} For what value of λ is the function f defined by \\ f (x) = \{\begin{cases} λ (x^{2} - 2 x), if x \leq 0 \\ 4 x +, if x > 0 \end{cases} \\ tinuous at x = 0 ? What about continuity at x = 1 ? \end{array}$

Ans.

$\begin{array}{l} The given function f is \\ f (x) = {\begin{cases} λ (x^{2} - 2 x), if x \leq 0 \\ 4 x +1, if x >0 \end{cases} \\ If f is continuous at x = 0, then \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \\ \Rightarrow \lim_{x \to 0^{-}} λ (x^{2} - 2 x) = \lim_{x \to 0^{+}} 4 x +1 = λ (0^{2} - 2 \times 0) \\ \Rightarrow λ (0^{2} - 2 \times 0) = 4 (0) + 1 = 0 \\ \Rightarrow 0 = 1 = 0, which is not possible . \\ Therefore, there is no value of λ for which f is continuous at x = 0 \\ At x = 1, \\ f (1) = 4 x + 1 = 4 \times 1 + 1 = 5 \\ \lim_{x \to 1} (4 x + 1) = 4 \times 1 + 1 = 5 \\ ∴ \lim_{x \to 1} f (x) = f (1) \\ Therefore, for any values of λ, f is continuous at x = 1 . \end{array}$

Q.19 Show that the function defined by g(x) = x – [x] is discontinuous at all integral
points. Here [x] denotes the greatest integer less than or equal to x.

Ans.

$\begin{array}{l} The given function is g (x) = x - [x] \\ Since, g is defined at all integral points . \\ Let n be an integer . \\ Then, \\ g (n) = n - [n] = n - n = 0 \\ The left hand limit of f at x = n is, \\ \lim_{x \to n^{-}} g (x) = \lim_{x \to n^{-}} (x - [x]) \\ = \lim_{x \to n^{-}} x - \lim_{x \to n^{-}} [x] \\ = n - (n - 1) \\ = 1 \\ The right hand limit of f at x = n is, \\ \lim_{x \to n^{+}} g (x) = \lim_{x \to n^{+}} (x - [x]) \\ = \lim_{x \to n^{+}} x - \lim_{x \to n^{+}} [x] \\ = n - n \\ = 0 \\ ∴ \lim_{x \to n^{-}} g (x) \neq \lim_{x \to n^{+}} g (x) \\ Therefore, f is not continuous at x = n . \\ Hence, g is discontinuous at all integral points . \end{array}$

Q.20

$\begin{array}{l} Is the function defined by f (x) = x^{2} - sinx + 5 continuous \\ at x = π ? \end{array}$

Ans.

$\begin{array}{l} The given function is f (x) = x^{2} - 6 sinx + 5 \\ âˆµ f is defined at x = π \\ At ‹ x = π, \\ f (π) = π^{2} - 6 sinπ + 5 \\ = π^{2} - 6 (0) + 5 \\ = π^{2} + 5 \\ \lim_{x \to π} f (x) = \lim_{x \to π} (x^{2} - 6 sinx + 5) \\ Put x \to π + h \\ If x \to π, then h \to 0 \\ ∴ \lim_{x \to π} f (x) = \lim_{x \to π} (x^{2} - 6 sinx + 5) \\ = \lim_{h \to 0} {{(h + π)}^{2} - 6 \sin (h + π) + 5} \\ = \lim_{h \to 0} {(h + π)}^{2} - 6 \lim_{h \to 0} \sin (h + π) + \lim_{h \to 0} 5 \\ = \lim_{h \to 0} {(h + π)}^{2} - 6 \lim_{h \to 0} (sinhcosπ + sinπcosh) + 5 \\ = {(0 + π)}^{2} - 6 \lim_{h \to 0} (\sinh \times - 1 + 0 \times \cosh) + 5 \\ = π^{2} + 6 \sin 0 + 5 \\ = π^{2} + 5 \\ ∴ \lim_{x \to π} f (x) = f (π) \\ Therefore, the given function f is continuous at x = π . \end{array}$

Q.21 Discuss the continuity of the following functions:
(a) f(x) = sin x + cos x
(b) f(x) = sin x − cos x
(c) f(x) = sin x . cos x

Ans.

$\begin{array}{l} If two functions (f and g) are continuous then their sum, \\ difference and product (i . e ., f + g, f - g and f . g) are also \\ continuous . \\ Let us prove that g (x) = sinx and h (x) = cosx are continuous \\ functions . \\ It is clear that g (x) = \sin x is defined for every real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ g (c) = sinc and \\ \lim_{x \to c} g (x) = \lim_{x \to c} sinx = \lim_{h \to 0} \sin (c + h) \\ = \lim_{h \to 0} (sinccosh + sinhcosc) \\ = sinccos 0 + \sin 0 cosc \\ = sinc \times 1 + 0 \times cosc \\ = sinc \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is a continuous function . \\ Let h (x) = \cos x \\ It is also evident that h (x) = \cos x is defined for every \\ real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ h (c) = cosc \\ \lim_{x \to c} h (x) = \lim_{x \to c} cosx = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosccosh - sinhsinc) \\ = cosccos 0 + \sin 0 sinc \\ = cosc \times 1 + 0 \times sinc \\ = cosc \\ ∴ \lim_{x \to c} h (x) = h (c) \\ Therefore, h is a continuous function . \\ Therefore, it can be concluded that \\ (a) f (x) = g (x) + h (x) = \sin x + \cos x is a continuous function \\ (b) f (x) = g (x) - h (x) = \sin x - \cos x is a continuous function \\ (c) f (x) = g (x) \times h (x) = \sin x \times \cos x is a continuous function \end{array}$

Q.22 Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Ans.

$\begin{array}{l} Firstly we have to prove that g (x) = sinx and h (x) = \cos x \\ are continuous functions . \\ It is clear that g (x) = \sin x is defined for every real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ g (c) = sinc and \\ \lim_{x \to c} g (x) = \lim_{x \to c} sinx = \lim_{h \to 0} \sin (c + h) \\ = \lim_{h \to 0} (sinccosh + sinhcosc) \\ = sinccos 0 + \sin 0 cosc \\ = sinc \times 1 + 0 \times cosc \\ = sinc \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is a continuous function . \\ Let h (x) = \cos x \\ It is also evident that h (x) = \cos x is defined for every \\ real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ h (c) = cosc \\ \lim_{x \to c} h (x) = \lim_{x \to c} cosx = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosccosh - sinhsinc) \\ = cosccos 0 + \sin 0 sinc \\ = cosc \times 1 + 0 \times sinc \\ = cosc \\ ∴ \lim_{x \to c} h (x) = h (c) \\ Therefore, h is a continuous function . \\ It can be concluded that, \\ cosecx = \frac{1}{sinx}, sinx \neq 0 is ‹ continuous . \\ \Rightarrow cosecx, nx \neq 0 (n \in Z) is ‹ continuous . \\ Therefore, cosecant is continuous except at x = nπ, n \in Z . \\ secx = \frac{1}{cosx}, cosx \neq 0 is ‹ continuous . \\ \Rightarrow secx, x \neq (2 n + 1) \frac{π}{2} (n \in Z) is ‹ continuous . \\ Therefore, secant is continuous except at x = (2 n + 1) \frac{π}{2} (n \in Z) . \\ cotx = \frac{cosx}{sinx}, sinx \neq 0 is ‹ continuous . \\ \Rightarrow cotx, x \neq nπ (n \in Z) is ‹ continuous . \\ Therefore, cotangent is continuous except at x = nπ, n \in Z . \end{array}$

Q.23

$\begin{array}{l} Find the points of discontinuity of f, where \\ f (x) = \{\begin{cases} \frac{sinx}{x}, if x < 0 \\ x +, if x \geq 0 \end{cases} \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} \frac{sinx}{x}, if x < 0 \\ x + 1, if x \geq 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c <0, then f (c) = \frac{sinc}{c} ‹‹ ‹ and \\ \lim_{x \to c} f (x) = \lim_{x \to c} \frac{sinx}{x} \\ = \frac{sinc}{c} \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x < 0 . \\ Case II : If c > 0, then f (c) = c + 1 \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x + 1) \\ = c + 1 \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x > 0 . \\ Case III : If c = 0, then \\ f (c) = f (0) \\ = 0 + 1 = 1 \\ The left hand limit of f at x = 0 is, \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} \frac{sinx}{x} = 1 \\ The right hand limit of f at x = 0 is, \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (x + 1) = 1 \\ ∴ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \\ Therefore, f is not continuous at x = 0 \\ From the above observations, it can be concluded that f is \\ continuous at all points of the real line . \\ Thus, f has no point of discontinuity . \end{array}$

Q.24

$\begin{array}{l} Determine if f defined by \\ f (x) = \{\begin{cases} x^{2} \sin \frac{}{x}, if x \neq 0 \\ , if x = 0 \end{cases} \\ is a continuous function ? \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} x^{2} \sin \frac{1}{x}, if x \neq 0 \\ 0, if x = 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ Case I : If c \neq 0, then f (c) = c^{2} \sin \frac{1}{c} ‹‹ ‹ and \\ \lim_{x \to c} f (x) = \lim_{x \to c} (x^{2} \sin \frac{1}{x}) \\ = c^{2} \sin \frac{1}{c} ‹‹ ‹ \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x \neq 0 . \\ Case II : If c = 0, then f (0) = 0 \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (x^{2} \sin \frac{1}{x}) \\ Since, - 1 \leq \sin \frac{1}{x} \leq 1, x \neq 0 \\ \Rightarrow - x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2} \\ \Rightarrow \lim_{x \to 0} - x^{2} \leq \lim_{x \to 0} x^{2} \sin \frac{1}{x} \leq \lim_{x \to 0} x^{2} \\ \Rightarrow 0 \leq \lim_{x \to 0} x^{2} \sin \frac{1}{x} \leq 0 \\ \Rightarrow \lim_{x \to 0} x^{2} \sin \frac{1}{x} = 0 \\ ∴ \lim_{x \to 0^{-}} f (x) = 0 \\ Similarly, \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (x^{2} \sin \frac{1}{x}) \\ = \lim_{x \to 0} x^{2} \sin \frac{1}{x} = 0 \\ ∴ \lim_{x \to 0^{-}} f (x) = f (0) = \lim_{x \to 0^{+}} f (x) \\ Therefore, f is continuous at x = 0 \\ From the above observations, it can be concluded that f is \\ continuous at every point of the real line . \\ Thus, f is a continuous function . \end{array}$

Q.25

$\begin{array}{l} Examine the continuity of f, where f is defined by \\ f (x) = \{\begin{cases} sinx - cosx, if x \neq 0 \\ -, if x = 0 \end{cases} \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} sinx - cosx, if x \neq 0 \\ - 1, if x = 1 \end{cases} \\ The given function f is defined at all the points of the real line . \\ Let c be a point on the real line . \\ CaseI : If c \neq 0, then f (c) = sinc - cosc \\ \lim_{x \to c} f (x) = \lim_{x \to c} (sinx - cosx) = sinc - cosc \\ ∴ \lim_{x \to c} f (x) = f (c) \\ Therefore, f is continuous at all points x, such that x \neq 0. \\ CaseII : \\ If c =0, then f (0) = - 1 \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (sinx - cosx) \\ = \sin 0 - \cos 0 \\ = 0 - 1 = - 1 \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (sinx - cosx) \\ = \sin 0 - \cos 0 \\ = 0 - 1 = - 1 \\ ∴ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \\ Therefore, f is continuous at x = 0 \\ From the above observations, it can be concluded that f is \\ continuous at every point of the real line . \\ Thus, f is a continuous function . \end{array}$

Q.26 Find the values of k so that the function f is continuous at the indicated point

$f(x)={\begin{array}{l} \frac{k cosx}{π - 2 x}, if x \neq \frac{π}{2} \\ 3, if x = 0 \end{array} atx = \frac{π}{2}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} \frac{kcosx}{π - 2 x}, if x \neq \frac{π}{2} \\ 3, if x = \frac{π}{2} \end{cases} \\ The given function f is continuous at x = \frac{π}{2}, if f is defined at \\ x = \frac{π}{2} and if the value of the f at x = \frac{π}{2} equals the limit of f at x = \frac{π}{2} . \\ Since, f is defined at x = \frac{π}{2} and f (\frac{π}{2}) = 3 \\ \lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{kcosx}{π - 2 x} \\ Let x = \frac{π}{2} + h, then x \to \frac{π}{2} \Rightarrow h \to 0 \\ ∴ \lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{kcosx}{π - 2 x} \\ = \lim_{h \to 0} \frac{kcos (\frac{π}{2} + h)}{π - 2 (\frac{π}{2} + h)} \\ = \lim_{h \to 0} \frac{- ksinh}{- 2 h} \\ = k \lim_{h \to 0} \frac{\sinh}{2 h} \\ = k \times \frac{1}{2} \\ = \frac{k}{2} \\ ∴ \lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2}) \\ \Rightarrow \frac{k}{2} = 3 \\ \Rightarrow k = 6 \\ Therefore, the required value of k is 6 . \end{array}$

Q.27

$\begin{array}{l} Find the values of k so that the function f is continuous at the indicated point . \\ f (x) = \{\begin{cases} x^{2}, if x \leq \\ , if x > 2 \end{cases} at x =2 \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} {kx}^{2}, if x \leq 2 \\ 3, if x > 2 \end{cases} \\ The given function f is continuous at x = 2, if f is defined at \\ x =2 and if the value of f at x =2 equals the limit of f at x =2 . \\ It is clear that f is defined at x = 2 and f (2) = k {(2)}^{2} = 4 k \\ \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{+}} f (x) = f (2) \\ \Rightarrow \lim_{x \to 2^{-}} ({kx}^{2}) = \lim_{x \to 2^{+}} (3) = 4 k \\ \Rightarrow k \times 2^{2} = 3 = 4 k \\ \Rightarrow 4 k = 3 \\ \Rightarrow k = \frac{3}{4} \\ Therefore, the required value of k is \frac{3}{4} . \end{array}$

Q.28 Find the values of k so that the function f is continuous at the indicated point

$f(x)={\begin{array}{l} k x + 1, if x \leq π \\ cosx, if x > π \end{array} atx = π$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} kx + 1, if x \leq π \\ cosx, if x > π \end{cases} \\ The given function f is continuous at x = π, then \\ L . H . L . = R . H . L . = f (π) \\ It is clear that f is defined at x = π and \\ f (π) = k (π) + 1 = πk + 1 \\ \lim_{x \to π^{-}} f (x) = \lim_{x \to π^{+}} f (x) = f (π) \\ \Rightarrow \lim_{x \to π^{-}} (kx + 1) = \lim_{x \to π^{+}} (cosx) = 4 π + 1 \\ \Rightarrow kπ + 1 = cosπ = 4 π + 1 \\ \Rightarrow kπ + 1 = - 1 = 4 π + 1 \\ \Rightarrow k = - \frac{2}{π} \\ Therefore, the required value of k is - \frac{2}{π} . \end{array}$

Q.29 Find the values of k so that the function f is continuous at the indicated point

$f(x)={\begin{array}{l} k x + 1, i f x \leq 5 \\ 3 x - 5, i f x > 5 \end{array} a t x = 5$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} kx + 1, if x \leq 5 \\ 3 x - 5, if x > 5 \end{cases} \\ The given function f is continuous at x = 5, then \\ L . H . L . = R . H . L . = f (π) \\ It is clear that f is defined at x = 5 and \\ f (5) = 5 k + 1 \\ \lim_{x \to 5^{-}} f (x) = \lim_{x \to 5^{+}} f (x) = f (5) \\ \Rightarrow \lim_{x \to 5^{-}} (kx + 1) = \lim_{x \to 5^{+}} (3 x - 5) = 5 k + 1 \\ \Rightarrow 5 k + 1 = 15 - 5 = 5 k + 1 \\ \Rightarrow 5 k + 1 = 10 \\ \Rightarrow k = \frac{9}{5} \\ Therefore, the required value of k is \frac{9}{5} . \end{array}$

Q.30

$\begin{array}{l} Find the values of a and b such that the function defined by \\ f (x) = \{\begin{cases} , if x \leq \\ ax + b, if < x < \\ , if x \geq \end{cases} \\ is a continuous function . \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = {\begin{cases} 5, if x \leq 2 \\ a x + b, if 2 < x < 10 \\ 21, i f x \geq 10 \end{cases} \\ It is evident that the given function f is defined at all points \\ of the real line .If f is a continuous function, then f is continuous \\ at all real numbers . \\ S o, f is continuous at x = 2 and x = 10 . \\ Since f is continuous at x = 2, we obtain \\ \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{+}} f (x) = f (2) \\ \Rightarrow \lim_{x \to 2^{-}} (5) = \lim_{x \to 2^{+}} (a x + b) = 5 \\ \Rightarrow 5 = 2 a + b = 5 \\ \Rightarrow 2 a + b = 5 … (i) \\ Since f is continuous at x = 10, we obtain \\ \lim_{x \to 10^{-}} f (x) = \lim_{x \to 10^{+}} f (x) = f (10) \\ \Rightarrow \lim_{x \to 10^{-}} (a x + b) = \lim_{x \to 2^{+}} (21) = 21 \\ \Rightarrow 10 a + b = 21 = 21 \\ \Rightarrow 10 a + b = 21 … (i i) \\ On subtracting equation (i) from equation (ii), we obtain \\ 8a = 16 \Rightarrow a = \frac{16}{8} = 2 \\ B y putting a = 2 in equation (ii), we get \\ 10 (2) + b = 21 \\ \Rightarrow b = 21 - 20 = 1 \\ Therefore, the values of a and b for which f is a continuous \\ function are 2 and 1 respectively . \end{array}$

Q.31 Show that the function defined by f(x) = cos(x²) is a continuous function.

Ans.

$\begin{array}{l} The given function is f (x) = \cos (x^{2}) \\ This function f is defined for every real number and f can be \\ written as the composition of two functions as, \\ f = gοh, where g (x) = \cos x and h (x) = x^{2} \\ [âˆµ (gοh) (x) = g (h (x)) = g (x^{2}) = {cosx}^{2}] \\ First we have to prove that g (x) = \cos x and h (x) = x^{2} are \\ continuous functions . \\ It is evident that g is defined for every real number . \\ Let c be a real number . \\ Then, g (c) = \cos c \\ Put x = c + h \\ If x \to c, then h \to 0 \\ \lim_{x \to c} g (x) = \lim_{x \to c} cosx \\ = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosccosh - sincsinh) \\ = \lim_{h \to 0} cosccosh - \lim_{h \to 0} sincsinh \\ = cosccos 0 - sincsin 0 \\ = cosc \times 1 - sinc \times 0 \\ = cosc \\ ∴ \lim_{x \to c} g (x) = cosc \\ Therefore, g (x) = \cos x is continuous function . \\ h (x) = x^{2} \\ Clearly, h is defined for every real number . \\ Let k be a real number, then h (k) = k^{2} \\ \lim_{x \to c} h (x) = \lim_{x \to c} x^{2} = k^{2} \\ \lim_{x \to c} h (x) = h (k) \\ Therefore, h is a continuous function . \\ It is known that for real valued functions g and h, such that \\ (goh) is defined at c, if g is continuous at c and if f is continuous \\ at g (c), then (f o g) is continuous at c . \\ Therefore, f (x) = (gοh) (x) = \cos (x^{2}) is a continuous function . \end{array}$

Q.32 Show that the function defined by f(x) = |cos x| is a continuous function.

Ans.

$\begin{array}{l} The given function is f (x) = | cosx | \\ This function f is defined for every real number and f can be \\ written as the composition of two functions as, \\ f = gοh, where g (x) = | x | and h (x) = cosx \\ [âˆµ (gοh) (x) = g (h (x)) = g (cosx) = | cosx |] \\ First we have to prove that g (x) = | x | and h (x) = cosx are \\ continuous functions . \\ g (x) = | x | = {\begin{cases} x, if x < 0 \\ - x, if x \geq 0 \end{cases} \\ Clearly, g is defined for all real numbers . \\ Let c be a real number . \\ CaseI : \\ If c <0, then g (c) = - c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (- x) = - c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x < 0 . \\ CaseII : \\ If c >0, then g (c) = c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (x) = c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x > 0 . \\ CaseIII : \\ If c = 0, then g (c) = g (0) = 0 \\ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{-}} (- x) = 0 \\ \lim_{x \to 0^{+}} g (x) = \lim_{x \to 0^{+}} (x) = 0 \\ ∴ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{+}} g (x) = g (0) \\ Therefore, g is continuous at x = 0 \\ From the above three observations, it can be concluded \\ that g is continuous at all points . \\ Since, h (x) = \cos x is defined for every real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ h (c) = \cos c \\ \lim_{x \to c} h (x) = \lim_{x \to c} cosx \\ = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosccosh - sincsinh) \\ = \lim_{h \to 0} cosccosh - \lim_{h \to 0} sincsinh \\ = cosccos 0 - sincsin 0 \\ = cosc \times 1 - sinc \times 0 \\ = cosc \\ ∴ \lim_{x \to c} h (x) = cosc \\ Therefore, h (x) = \cos x is a continuous function . \\ It is known that for real valued functions g and h, such that \\ (gοh) is defined at c, if g is continuous at c and if f is continuous \\ at g (c), then (f o g) is continuous at c . \\ Therefore, f (x) = (gοh) (x) \\ = g (h (x)) \\ = g (cosx) = | cosx | is continuous function . \end{array}$

Q.33 Examine that sin|x| is a continuous function.

Ans.

$\begin{array}{l} The given function is f (x) = \sin | x | \\ This function f is defined for every real number and f can be \\ written as the composition of two functions as, \\ f = gοh, where g (x) = | x | and h (x) = sinx \\ [âˆµ (gοh) (x) = g (h (x)) = g (| x |) = \sin | x | = f (x)] \\ First we have to prove that g (x) = | x | and h (x) = sinx are \\ continuous functions . \\ g (x) = | x | = {\begin{cases} x, if x < 0 \\ - x, if x \geq 0 \end{cases} \\ Clearly, g is defined for all real numbers . \\ Let c be a real number . \\ CaseI : \\ If c <0, then g (c) = - c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (- x) = - c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x < 0 . \\ CaseII : \\ If c >0, then g (c) = c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (x) = c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x > 0 . \\ CaseIII : \\ If c = 0, then g (c) = g (0) = 0 \\ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{-}} (- x) = 0 \\ \lim_{x \to 0^{+}} g (x) = \lim_{x \to 0^{+}} (x) = 0 \\ ∴ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{+}} g (x) = g (0) \\ Therefore, g is continuous at x = 0 \\ From the above three observations, it can be concluded \\ that g is continuous at all points . \\ Since, h (x) = sinx is defined for every real number . \\ Let c be a real number . Put x = c + h \\ If x \to c, then h \to 0 \\ h (c) = sinc \\ \lim_{x \to c} h (x) = \lim_{x \to c} sinx \\ = \lim_{h \to 0} \sin (c + h) \\ = \lim_{h \to 0} (sinccosh - coscsinh) \\ = \lim_{h \to 0} sinccosh - \lim_{h \to 0} coscsinh \\ = sinccos 0 - coscsin 0 \\ = sinc \times 1 - cosc \times 0 \\ = sinc \\ ∴ \lim_{x \to c} h (x) = sinc \\ Therefore, h (x) is a continuous function . \\ It is known that for real valued functions g and h, such that \\ (gοh) is defined at c, if g is continuous at c and if f is continuous \\ at g (c), then (f o g) is continuous at c . \\ Therefore, f (x) = (gοh) (x) \\ = g (h (x)) \\ = g (sinx) = | sinx | is continuous function . \end{array}$

Q.34 Find all the points of discontinuity of f defined by f(x) = |x| – |x+1|.

Ans.

$\begin{array}{l} The given function is f (x) = | x | - | x + 1 | \\ The two functions, g and h, are defined as \\ g (x) = | x | and h (x) = | x + 1 | \\ Then, f = g - h \\ The continuity of g and h is checked first . \\ g (x) = | x | = {\begin{cases} x, if x < 0 \\ - x, if x \geq 0 \end{cases} \\ Clearly, g is defined for all real numbers . \\ Let c be a real number . \\ CaseI : \\ If c <0, then g (c) = - c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (- x) = - c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x < 0 . \\ CaseII : \\ If c >0, then g (c) = c \\ and \lim_{x \to c} g (x) = \lim_{x \to c} (x) = c \\ ∴ \lim_{x \to c} g (x) = g (c) \\ Therefore, g is continuous at all points x, such that x > 0 . \\ CaseIII : \\ If c = 0, then g (c) = g (0) = 0 \\ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{-}} (- x) = 0 \\ \lim_{x \to 0^{+}} g (x) = \lim_{x \to 0^{+}} (x) = 0 \\ ∴ \lim_{x \to 0^{-}} g (x) = \lim_{x \to 0^{+}} g (x) = g (0) \\ Therefore, g is continuous at x = 0 \\ From the above three observations, it can be concluded \\ that g is continuous at all points . \\ h (x) = | x + 1 | can be written as \\ h (x) = {\begin{cases} - (x + 1), if x < - 1 \\ x + 1, if x \geq - 1 \end{cases} \\ Since, h is defined for every real number . \\ Let c be a real number . \\ CaseI : \\ If c < - 1, then h (c) = - (c + 1) \\ and \lim_{x \to c} h (x) = \lim_{x \to c} [- (x + 1)] \\ = - (c + 1) \\ ∴ \lim_{x \to c} h (x) = h (c) \\ Therefore, h is continuous at all points x, such that x < - 1. \\ CaseII : \\ Ifc > - 1, thenh (c) = c + 1 and \\ \lim_{x \to c} h (x) = \lim_{x \to c} [(x + 1)] \\ = (c + 1) \\ ∴ \lim_{x \to c} h (x) = h (c) \\ Therefore, h is continuous at all points x, such that x > - 1 \\ CaseIII : \\ Ifc = - 1, thenh (c) = h (- 1) = - 1 + 1 = 0 \\ \lim_{x \to - 1^{-}} h (x) = \lim_{x \to - 1^{-}} [- (x + 1)] = - (- 1 + 1) = 0 \\ \lim_{x \to - 1^{+}} h (x) = \lim_{x \to - 1^{+}} [(x + 1)] = (- 1 + 1) = 0 \\ ∴ \lim_{x \to - 1^{-}} h (x) = \lim_{x \to - 1^{+}} h (x) = h (- 1) \\ Therefore, h is continuous at x = - 1 \\ From the above three observations, it can be concluded \\ that h is continuous at all points of the real line . \\ g and h are continuous functions . Therefore, f = g - h \\ is also a continuous function . \\ Therefore, f has no point of discontinuity . \end{array}$

Q.35 Differentiate the function with respect to x.

sin(x² + 5)

Ans.

$\begin{array}{l} Let ‹ f (x) = \sin (x^{2} + 5), u (x) = x^{2} + 5 and v (t) = sint . Then \\ (vοu) (x) = v (u (x)) = v (x^{2} + 5) = \sin (x^{2} + 5) = f (x) \\ Thus, f is composite of two functions . \\ Put t = u (x) = x^{2} + 5, then \\ \frac{dv}{dt} = \frac{d}{dt} sint = cost = \cos (x^{2} + 5) \\ \frac{dt}{dx} = \frac{d}{dx} (x^{2} + 5) = 2 x + 0 = 2 x \\ Hence, by chain rule \\ \frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} \\ = \cos (x^{2} + 5) . 2 x \\ = 2 xcos (x^{2} + 5) \\ ∴ \frac{d}{dx} \sin (x^{2} + 5) = 2 xcos (x^{2} + 5) \end{array}$

Q.36 Differentiate the function with respect to x.

cos(sinx)

Ans.

$\begin{array}{l} Let y = \cos (sinx) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} [\cos (sinx)] \\ = - \sin (sinx) . \frac{d}{dx} sinx [By chain rule] \\ = - \sin (sinx) . cosx \\ = - cosx . \sin (sinx) \\ ∴ \frac{d}{dx} \cos (sinx) = - cosx . \sin (sinx) \end{array}$

Q.37 Differentiate the function with respect to x.

sin (ax+b)

Ans.

$\begin{array}{l} Let y = \sin (ax + b) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} [\sin (ax + b)] \\ = \cos (ax + b) . \frac{d}{dx} (ax + b) [By chain rule] \\ = \cos (ax + b) . a \\ = acos (ax + b) \\ ∴ \frac{d}{dx} \sin (ax + b) = acos (ax + b) \end{array}$

Q.38 Differentiate the function with respect to x.

$sec (tan (\sqrt{x}))$

Ans.

$\begin{array}{l} Let y = \sec (\tan (\sqrt{x})) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \sec (\tan (\sqrt{x})) \\ = \sec (\tan (\sqrt{x})) \tan (\tan (\sqrt{x})) \frac{d}{dx} \tan (\sqrt{x}) \\ [âˆµ \frac{d}{dx} secx = secxtanx] \\ = \sec (\tan (\sqrt{x})) \tan (\tan (\sqrt{x})) . \sec^{2} (\sqrt{x}) \frac{d}{dx} (\sqrt{x}) \\ [âˆµ \frac{d}{dx} tanx = \sec^{2} x] \\ = \sec (\tan (\sqrt{x})) \tan (\tan (\sqrt{x})) . \sec^{2} (\sqrt{x}) . \frac{1}{2 \sqrt{x}} \\ ∴ \frac{d}{dx} \sec (\tan (\sqrt{x})) = \frac{1}{2 \sqrt{x}} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) . \sec^{2} (\sqrt{x}) \end{array}$

Q.39

$Differentiate the functions with respect to x \frac{s i n (a x + b)}{c o s (c x + d)}$

Ans.

$\begin{array}{l} Let y = \frac{\sin (ax + b)}{\cos (cx + d)} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} {\frac{\sin (ax + b)}{\cos (cx + d)}} \\ = \frac{\cos (cx + d) \frac{d}{dx} \sin (ax + b) - \sin (ax + b) \frac{d}{dx} \cos (cx + d)}{\cos^{2} (cx + d)} \\ [By Quotient rule] \\ = \frac{\cos (cx + d) \cos (ax + b) \frac{d}{dx} (ax + b) - \sin (ax + b) \times - \sin (cx + d) \frac{d}{dx} (cx + d)}{\cos^{2} (cx + d)} \\ = \frac{acos (cx + d) \cos (ax + b) + csin (ax + b) \sin (cx + d)}{\cos^{2} (cx + d)} \\ = \frac{acos (ax + b)}{\cos (cx + d)} + \frac{csin (ax + b) \sin (cx + d)}{\cos^{2} (cx + d)} \\ = acos (ax + b) \sec (cx + d) + csin (ax + b) \tan (cx + d) \sec (cx + d) \\ ∴ \frac{d}{dx} \frac{\sin (ax + b)}{\cos (cx + d)} = acos (ax + b) \sec (cx + d) \\ + csin (ax + b) \tan (cx + d) \sec (cx + d) \end{array}$

Q.40 Differentiate the function with respect to x.

cosx³. sin²(x⁵)

Ans.

$\begin{array}{l} Let y = {cosx}^{3} . \sin^{2} (x^{5}) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = {cosx}^{3} \frac{d}{dx} \sin^{2} (x^{5}) + \sin^{2} (x^{5}) \frac{d}{dx} {cosx}^{3} [By ‹ Product Rule .] \\ = {cosx}^{3} \frac{d}{dx} {\sin (x^{5})}^{2} + \sin^{2} (x^{5}) \times - {sinx}^{3} \frac{d}{dx} x^{3} \\ = {cosx}^{3} \times 2 {sinx}^{5} \frac{d}{dx} \sin (x^{5}) - \sin^{2} (x^{5}) {sinx}^{3} \times 3 x^{2} \\ = 2 {cosx}^{3} {sinx}^{5} \times \cos (x^{5}) \frac{d}{dx} x^{5} - 3 x^{2} \sin^{2} (x^{5}) {sinx}^{3} \\ = 2 {cosx}^{3} {sinx}^{5} \cos (x^{5}) \times 5 x^{4} - 3 x^{2} \sin^{2} (x^{5}) {sinx}^{3} \\ = 10 x^{4} {sinx}^{5} \cos (x^{5}) {cosx}^{3} - 3 x^{2} {sinx}^{3} \sin^{2} (x^{5}) \\ ∴ \frac{d}{dx} {cosx}^{3} . \sin^{2} (x^{5}) = 10 x^{4} {sinx}^{5} \cos (x^{5}) {cosx}^{3} - 3 x^{2} {sinx}^{3} \sin^{2} (x^{5}) \end{array}$

Q.41

$Differentiate the function with respect to x. cos (\sqrt{x})$

Ans.

$\begin{array}{l} Let y = \cos (\sqrt{x}) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \cos (\sqrt{x}) \\ = - \sin \sqrt{x} \frac{d}{dx} \sqrt{x} \\ = - \sin \sqrt{x} \times \frac{1}{2} x^{- \frac{1}{2}} \\ = - \frac{\sin \sqrt{x}}{2 \sqrt{x}} \end{array}$

Q.42

$Differentiate the function with respect to x . 2 \sqrt{\cot (x^{2})}$

Ans.

$\begin{array}{l} Let y = 2 \sqrt{\cot (x^{2})} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = 2 \frac{d}{dx} \sqrt{\cot (x^{2})} \\ = 2 \times \frac{1}{2} {\cot (x^{2})}^{- \frac{1}{2}} \frac{d}{dx} \cot (x^{2}) [By ‹ Chain Rule] \\ = {\cot (x^{2})}^{- \frac{1}{2}} \times - {cosec}^{2} (x^{2}) \frac{d}{dx} x^{2} \\ = {\cot (x^{2})}^{- \frac{1}{2}} \times - {cosec}^{2} (x^{2}) \times 2 x \\ = - \frac{2 {xcosec}^{2} (x^{2})}{\sqrt{\cot (x^{2})}} \\ = - \frac{2 x}{\sin^{2} x^{2} \sqrt{\frac{{cosx}^{2}}{{sinx}^{2}}}} \\ = - \frac{2 x}{{sinx}^{2} \sqrt{{sinx}^{2} {cosx}^{2}}} \\ = - \frac{2 \sqrt{2} x}{{sinx}^{2} \sqrt{2 {sinx}^{2} {cosx}^{2}}} \\ = - \frac{2 \sqrt{2} x}{{sinx}^{2} \sqrt{\sin 2 x^{2}}} \\ ∴ \frac{d}{dx} 2 \sqrt{\cot (x^{2})} = \frac{- 2 \sqrt{2} x}{{sinx}^{2} \sqrt{\sin 2 x^{2}}} \end{array}$

Q.43

$\begin{array}{l} Prove that the function f given by f (x) = |x - 1|, x \in is not \\ differentiable at x = 1 . \end{array}$

Ans.

$\begin{array}{l} The given function is f (x) = | x - 1 |, x \in R \\ Since a function f is differentiable at a point c in its domain if both \\ \lim_{x \to 0^{-}} \frac{f (c + h) - f (c)}{h} and \lim_{x \to 0^{+}} \frac{f (c + h) - f (c)}{h} are finite and equal . \\ Now, left hand limit at x = 1, \\ \lim_{h \to 0^{-}} \frac{f (1 + h) - f (1)}{h} = \lim_{h \to 0^{-}} \frac{| 1 + h - 1 | - | 1 - 1 |}{h} \\ = \lim_{h \to 0^{-}} \frac{| h |}{h} \\ = \lim_{h \to 0^{-}} \frac{- h}{h} (h < 0 \Rightarrow | h | = - h) \\ = - 1 \\ Now, right hand limit at x = 1, \\ \lim_{h \to 0^{+}} \frac{f (1 + h) - f (1)}{h} = \lim_{h \to 0^{+}} \frac{| 1 + h - 1 | - | 1 - 1 |}{h} \\ = \lim_{h \to 0^{-}} \frac{| h |}{h} \\ = \lim_{h \to 0^{-}} \frac{h}{h} (h > 0 \Rightarrow | h | = h) \\ = 1 \\ Since the left and right hand limits of f at x = 1 are not equal, \\ f is not differentiable at x = 1 . \end{array}$

Q.44

$\begin{array}{l} Prove that the greatest integer function defined ‹ by \\ f (x) = [x], 0 < x < 3 \\ is not differentiable at ‹ x = 1 and x = 2 . \end{array}$

Ans.

$\begin{array}{l} The given function f is f (x) = [x], 0 < x < 3 \\ Since a function f is differentiable at a point c in its domain if both \\ \lim_{x \to 0^{-}} \frac{f (c + h) - f (c)}{h} and \lim_{x \to 0^{+}} \frac{f (c + h) - f (c)}{h} are finite and equal . \\ Now, left hand limit at x = 1, \\ \lim_{h \to 0^{-}} \frac{f (1 + h) - f (1)}{h} = \lim_{h \to 0^{-}} \frac{[1 + h] - [1]}{h} \\ = \lim_{h \to 0^{-}} \frac{0 - 1}{h} \\ = \lim_{h \to 0^{-}} \frac{- 1}{h} = \infty \\ And, left hand limit at x = 1, \\ \lim_{h \to 0^{+}} \frac{f (1 + h) - f (1)}{h} = \lim_{h \to 0^{+}} \frac{[1 + h] - [1]}{h} \\ = \lim_{h \to 0^{+}} \frac{1 - 1}{h} \\ = \lim_{h \to 0^{+}} \frac{0}{h} = 0 \\ Since the left and right hand limits of f at x = 1 are not equal, \\ f is not differentiable at x = 1 \\ Checking the differentiability of the given function at x = 2, \\ So, left hand limit at x = 2, \\ \lim_{h \to 0^{-}} \frac{f (2 + h) - f (2)}{h} = \lim_{h \to 0^{-}} \frac{[2 + h] - [2]}{h} \\ = \lim_{h \to 0^{-}} \frac{1 - 2}{h} \\ = \lim_{h \to 0^{-}} \frac{- 1}{h} = \infty \\ And, left hand limit at x = 2, \\ \lim_{h \to 0^{+}} \frac{f (2 + h) - f (2)}{h} = \lim_{h \to 0^{+}} \frac{[2 + h] - [2]}{h} \\ = \lim_{h \to 0^{+}} \frac{2 - 2}{h} \\ = \lim_{h \to 0^{+}} \frac{0}{h} = 0 \\ Since the left and right hand limits of f at x = 2 are not equal, \\ f is not differentiable at x = 2 . \end{array}$

Q.45 Find dy/dx in the following:
2x + 3y = sinx

Ans.

$\begin{array}{l} Given : 2 x + 3 y = sinx \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (2 x + 3 y) = \frac{d}{dx} sinx \\ 2 \frac{d}{dx} x + 3 \frac{dy}{dx} = cosx \\ 2 + 3 \frac{dy}{dx} = cosx \\ \frac{dy}{dx} = \frac{cosx - 2}{3} \end{array}$

Q.46 Find dy/dx in the following:
2x + 3y = siny

Ans.

$\begin{array}{l} Given : 2 x + 3 y = siny \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (2 x + 3 y) = \frac{d}{dx} siny \\ 2 \frac{d}{dx} x + 3 \frac{dy}{dx} = cosy \frac{dy}{dx} \\ 2 + 3 \frac{dy}{dx} = cosy \frac{dy}{dx} \\ 3 \frac{dy}{dx} - cosy \frac{dy}{dx} = - 2 \\ \frac{dy}{dx} (3 - cosy) = - 2 \\ \frac{dy}{dx} = \frac{- 2}{(3 - cosy)} \\ ∴ \frac{dy}{dx} = \frac{2}{(cosy - 3)} \end{array}$

Q.47

$Find \frac{dy}{dx} in the following : ax + {by}^{2} = cosy$

Ans.

$\begin{array}{l} Given : ax + {by}^{2} = cosy \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (ax + {by}^{2}) = \frac{d}{dx} cosy \\ a \frac{d}{dx} x + 2 by \frac{dy}{dx} = - siny \frac{dy}{dx} \\ a + 2 by \frac{dy}{dx} = - siny \frac{dy}{dx} \\ 2 by \frac{dy}{dx} + siny \frac{dy}{dx} = - a \\ \frac{dy}{dx} (2 by + siny) = - a \\ \frac{dy}{dx} = \frac{- a}{(2 by + siny)} \\ ∴ \frac{dy}{dx} = \frac{- a}{(2 by + siny)} \end{array}$

Q.48

$Find \frac{dy}{dx} in the following : xy + y^{2} = tanx + y$

Ans.

$\begin{array}{l} Given : xy + y^{2} = tanx + y \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (xy + y^{2}) = \frac{d}{dx} (tanx + y) \\ [x \frac{dy}{dx} + y \frac{d}{dx} x] + \frac{d}{dx} y^{2} = \sec^{2} x + \frac{dy}{dx} [By ‹ Product Rule] \\ x \frac{dy}{dx} + y + 2 y \frac{dy}{dx} = \sec^{2} x + \frac{dy}{dx} \\ x \frac{dy}{dx} + 2 y \frac{dy}{dx} - \frac{dy}{dx} = \sec^{2} x - y \\ (x + 2 y - 1) \frac{dy}{dx} = \sec^{2} x - y \\ \frac{dy}{dx} = \frac{\sec^{2} x - y}{(x + 2 y - 1)} \end{array}$

Q.49

$Find \frac{dy}{dx} in the following x^{3} + x^{2} y + {xy}^{2} + y^{3} = 81$

Ans.

$\begin{array}{l} Given : x^{3} + x^{2} y + {xy}^{2} + y^{3} = 81 \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (x^{3} + x^{2} y + {xy}^{2} + y^{3}) = \frac{d}{dx} 81 \\ \frac{d}{dx} x^{3} + \frac{d}{dx} x^{2} y + \frac{d}{dx} {xy}^{2} + \frac{d}{dx} y^{3} = 0 \\ 3 x^{2} + (x^{2} \frac{dy}{dx} + y \frac{d}{dx} x^{2}) + (x \frac{d}{dx} y^{2} + y^{2} \frac{d}{dx} x) + 3 y^{2} \frac{dy}{dx} = 0 \\ [By product rule] \\ 3 x^{2} + x^{2} \frac{dy}{dx} + y \times 2 x + (2 xy \frac{dy}{dx} + y^{2} \times 1) + 3 y^{2} \frac{dy}{dx} = 0 \\ 3 x^{2} + x^{2} \frac{dy}{dx} + 2 xy + 2 xy \frac{dy}{dx} + y^{2} + 3 y^{2} \frac{dy}{dx} = 0 \\ (x^{2} + 2 xy + 3 y^{2}) \frac{dy}{dx} = - 3 x^{2} - 2 xy - y^{2} \\ ∴ \frac{dy}{dx} = - \frac{(3 x^{2} + 2 xy + y^{2})}{(x^{2} + 2 xy + 3 y^{2})} \end{array}$

Q.50

$\begin{array}{l} Given : x^{2} + xy + y^{2} = 100 \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (x^{2} + xy + y^{2}) = \frac{d}{dx} 100 \\ \frac{d}{dx} x^{2} + \frac{d}{dx} xy + \frac{d}{dx} y^{2} = 0 \\ 2 x + (x \frac{dy}{dx} + y \frac{d}{dx} x) + 2 y \frac{dy}{dx} = 0 [By product rule] \\ 2 x + x \frac{dy}{dx} + y \times 1 + 2 y \frac{dy}{dx} = 0 \\ \frac{dy}{dx} (x + 2 y) = - 2 x - y \\ \frac{dy}{dx} = - \frac{2 x + y}{x + 2 y} \end{array}$

Ans.

$\begin{array}{l} Find \frac{dy}{dx} in the following \\ x^{2} + xy + y^{2} = 100 \end{array}$

Q.51

$\begin{array}{l} Find \frac{dy}{dx} in the following \\ x^{2} + xy + y^{2} = 100 \end{array}$

Ans.

Q.52

${sin}^{2} y + cosxy = K$

Ans.

$\begin{array}{l} Given : \sin^{2} y + cosxy = K \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (\sin^{2} y + cosxy) = \frac{d}{dx} K \\ \frac{d}{dx} \sin^{2} y + \frac{d}{dx} cosxy = 0 \\ 2 siny \frac{d}{dx} siny - sinxy \frac{d}{dx} (xy) = 0 [Using chain rule] \\ 2 sinycosy \frac{dy}{dx} - sinxy (x \frac{dy}{dx} + y \frac{d}{dx} x) = 0 [By product rule] \\ \sin 2 y \frac{dy}{dx} - xsinxy \frac{dy}{dx} - ysinxy = 0 \\ \frac{dy}{dx} (\sin 2 y - xsinxy) = ysinxy \\ \frac{dy}{dx} = \frac{ysinxy}{(\sin 2 y - xsinxy)} \end{array}$

Q.53

$\begin{array}{l} Find \frac{dy}{dx} in the following \\ \sin^{2} x + \cos^{2} y = 1 \end{array}$

Ans.

$\begin{array}{l} Given : \sin^{2} x + \cos^{2} y = 1 \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} (\sin^{2} x + \cos^{2} y) = \frac{d}{dx} 1 \\ \Rightarrow \frac{d}{dx} \sin^{2} x + \frac{d}{dx} \cos^{2} y = 0 \\ \Rightarrow 2 sinx \frac{d}{dx} sinx + 2 cosy \frac{d}{dx} (cosy) = 0 [Using chain rule] \\ \Rightarrow 2 sinxcosx + 2 cosy (- siny) \frac{dy}{dx} = 0 \\ \Rightarrow \sin 2 x - \sin 2 y \frac{dy}{dx} = 0 \\ \Rightarrow - \sin 2 y \frac{dy}{dx} = - \sin 2 x \\ \Rightarrow \frac{dy}{dx} = \frac{\sin 2 x}{\sin 2 y} \end{array}$

Q.54

$Find \frac{dy}{dx} in the following y = \sin^{- 1} (\frac{2 x}{1 + x^{2}})$

Ans.

$\begin{array}{l} The given relationship is \\ y = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \\ \Rightarrow siny = \frac{2 x}{1 + x^{2}} \\ Differentiating both sides w . r . t . x, we get \\ \Rightarrow \frac{d}{dx} siny = \frac{d}{dx} (\frac{2 x}{1 + x^{2}}) \\ \Rightarrow cosy \frac{dy}{dx} = \frac{(1 + x^{2}) \frac{d}{dx} 2 x - 2 x \frac{d}{dx} (1 + x^{2})}{{(1 + x^{2})}^{2}} \\ [By Quotient Rule] \\ \Rightarrow \sqrt{1 - \sin^{2} y} \frac{dy}{dx} = \frac{(1 + x^{2}) \times 2 - 2 x (0 + 2 x)}{{(1 + x^{2})}^{2}} \\ \Rightarrow \sqrt{1 - {(\frac{2 x}{1 + x^{2}})}^{2}} \frac{dy}{dx} = \frac{2 + 2 x^{2} - 4 x^{2}}{{(1 + x^{2})}^{2}} [Putting value of siny] \\ \Rightarrow \sqrt{\frac{{(1 + x^{2})}^{2} - 4 x^{2}}{{(1 + x^{2})}^{2}}} \frac{dy}{dx} = \frac{2 - 2 x^{2}}{{(1 + x^{2})}^{2}} \\ \Rightarrow \sqrt{\frac{{(1 - x^{2})}^{2}}{{(1 + x^{2})}^{2}}} \frac{dy}{dx} = \frac{2 - 2 x^{2}}{{(1 + x^{2})}^{2}} \\ \Rightarrow \frac{(1 - x^{2})}{(1 + x^{2})} \frac{dy}{dx} = \frac{2 - 2 x^{2}}{{(1 + x^{2})}^{2}} \\ \Rightarrow \frac{dy}{dx} = \frac{2 - 2 x^{2}}{(1 + x^{2}) (1 - x^{2})} \\ \Rightarrow \frac{dy}{dx} = \frac{2 (1 - x^{2})}{(1 + x^{2}) (1 - x^{2})} \\ ∴ \frac{dy}{dx} = \frac{2}{(1 + x^{2})} \end{array}$

Q.55

$Find \frac{dy}{dx} in the following : y = \tan^{- 1} (\frac{3 x - x^{3}}{1 + 3 x^{2}}), - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$

Ans.

$\begin{array}{l} We ‹‹ have y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) \\ Putting x = tanθ, ‹ we get \\ y = \tan^{- 1} (\frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) \\ = \tan^{- 1} (\tan 3 θ) \\ Since, - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}, then \\ x = tanθ \\ \Rightarrow - \frac{1}{\sqrt{3}} < tanθ < \frac{1}{\sqrt{3}} \\ \Rightarrow - \frac{π}{6} < θ < \frac{π}{6} \\ \Rightarrow - \frac{π}{2} < 3 θ < \frac{π}{2} \\ ∴ y = 3 θ [âˆµ - \frac{π}{2} < 3 θ < \frac{π}{2}] \\ = 3 \tan^{- 1} x [âˆµ x = tanθ \Rightarrow θ = \tan^{- 1} x] \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = 3 \frac{d}{dx} \tan^{- 1} x \\ = 3 \times \frac{1}{1 + x^{2}} \\ = \frac{3}{1 + x^{2}} \\ ∴ \frac{d}{dx} \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) = \frac{3}{1 + x^{2}} \end{array}$

Q.56

$Find \frac{dy}{dx} in the following : y = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1$

Ans.

$\begin{array}{l} We ‹‹ have y = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1 \\ Putting x = tanθ, ‹ we get \\ y = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \\ = \cos^{- 1} (\cos 2 θ) \\ Since, 0 < x < 1, then \\ x = tanθ \\ \Rightarrow 0 < tanθ < 1 \\ \Rightarrow 0 < θ < \frac{π}{4} \\ \Rightarrow 0 < 2 θ < \frac{π}{2} \\ ∴ y = 2 θ [âˆµ 0 < 2 θ < \frac{π}{2}] \\ = 2 \tan^{- 1} x [âˆµ x = tanθ \Rightarrow θ = \tan^{- 1} x] \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = 2 \frac{d}{dx} \tan^{- 1} x \\ = 2 \times \frac{1}{1 + x^{2}} \\ = \frac{2}{1 + x^{2}} \\ ∴ \frac{d}{dx} \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) = \frac{2}{1 + x^{2}} \end{array}$

Q.57

Ans.

$\begin{array}{l} We ‹‹ have y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1 \\ Putting x = tanθ, ‹ we get \\ y = \sin^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \\ = \sin^{- 1} (\cos 2 θ) \\ = \sin^{- 1} {\sin (\frac{π}{2} - 2 θ)} \\ Since, 0 < x < 1, then \\ x = tanθ \\ \Rightarrow 0 < tanθ < 1 \\ \Rightarrow 0 < θ < \frac{π}{4} \\ \Rightarrow 0 < 2 θ < \frac{π}{2} \\ \Rightarrow 0 < \frac{π}{2} - 2 θ < \frac{π}{2} \\ ∴ y = \frac{π}{2} - 2 θ [âˆµ 0 < \frac{π}{2} - 2 θ < \frac{π}{2}] \\ = \frac{π}{2} - 2 \tan^{- 1} x [âˆµ x = tanθ \Rightarrow θ = \tan^{- 1} x] \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = 0 - 2 \frac{d}{dx} \tan^{- 1} x \\ = - 2 \times \frac{1}{1 + x^{2}} \\ = - \frac{2}{1 + x^{2}} \\ Thus, \\ \frac{d}{dx} \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = - \frac{2}{1 + x^{2}} \end{array}$

Q.58

$Find \frac{dy}{dx} in the following : y = \cos^{- 1} (\frac{2 x}{1 + x^{2}}), - 1 < x < 1$

Ans.

$\begin{array}{l} We ‹‹ have y = \cos^{- 1} (\frac{2 x}{1 + x^{2}}), - 1 < x < 1 \\ Putting x = tanθ, ‹ we get \\ y = \cos^{- 1} (\frac{2 tanθ}{1 + \tan^{2} θ}) \\ = \cos^{- 1} (\sin 2 θ) \\ = \cos^{- 1} {\cos (\frac{π}{2} - 2 θ)} \\ Since, - 1 < x < 1, then \\ x = tanθ \\ \Rightarrow - 1 < tanθ < 1 \\ \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} \\ \Rightarrow - \frac{π}{2} < 2 θ < \frac{π}{2} \\ \Rightarrow \frac{π}{2} > - 2 θ > - \frac{π}{2} \\ \Rightarrow π > \frac{π}{2} - 2 θ > 0 \\ \Rightarrow 0 < \frac{π}{2} - 2 θ < π \\ ∴ y = \frac{π}{2} - 2 θ \\ = \frac{π}{2} - 2 \tan^{- 1} x [âˆµ x = tanθ \Rightarrow θ = \tan^{- 1} x] \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \frac{π}{2} - 2 \frac{d}{dx} \tan^{- 1} x \\ = 0 - 2 \times \frac{1}{1 + x^{2}} \\ ∴ \frac{d}{dx} \cos^{- 1} (\frac{2 x}{1 + x^{2}}) \\ = - \frac{2}{1 + x^{2}} . \end{array}$

Q.59

$Find \frac{dy}{dx} in the following y = \sin^{- 1} (2 x \sqrt{1 - x^{2}}), - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

Ans.

$\begin{array}{l} We ‹‹ have y = \sin^{- 1} (2 x \sqrt{1 - x^{2}}), - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \\ Putting x = sinθ, ‹ we get \\ y = \sin^{- 1} (2 sinθ \sqrt{1 - \sin^{2} θ}) \\ = \sin^{- 1} (2 sinθcosθ) \\ = \sin^{- 1} (\sin 2 θ) \\ Since, - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}, then \\ x = sinθ \\ \Rightarrow - \frac{1}{\sqrt{2}} < sinθ < \frac{1}{\sqrt{2}} \\ \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} \\ \Rightarrow - \frac{π}{2} < 2 θ < \frac{π}{2} \\ ∴ y = 2 θ \\ = 2 \sin^{- 1} x [x = sinθ \Rightarrow θ = \sin^{- 1} x] \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (2 \sin^{- 1} x) \\ = \frac{2}{\sqrt{1 - x^{2}}} \\ Thus, \\ \frac{d}{dx} {\sin^{- 1} (2 x \sqrt{1 - x^{2}})} = \frac{2}{\sqrt{1 - x^{2}}} \end{array}$

Q.60 Find dy/dx in the following

${y = sec}^{- 1} (\frac{1}{2 x^{2} - 1}), 0 < x < \frac{1}{\sqrt{2}}$

Ans.

$\begin{array}{l} We have \\ y = \sec^{- 1} (\frac{1}{2 x^{2} - 1}), 0 < x < \frac{1}{\sqrt{2}} \\ = \cos^{- 1} (2 x^{2} - 1) \\ Putting x = cosθ, ‹ we get \\ y = \cos^{- 1} (2 \cos^{2} θ - 1) \\ = \cos^{- 1} (\cos 2 θ) \\ Since, 0 < x < \frac{1}{\sqrt{2}} \\ \Rightarrow 0 < cosθ < \frac{1}{\sqrt{2}} [âˆµ x = cosθ] \\ \Rightarrow \frac{3 π}{2} < θ < \frac{7 π}{4} \\ \Rightarrow 0 < θ < \frac{π}{4} \\ \Rightarrow 0 < 2 θ < \frac{π}{2} \\ ∴ y = 2 θ \\ = 2 \cos^{- 1} x [âˆµ x = cosθ \Rightarrow θ = \cos^{- 1} x] \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = 2 \frac{d}{dx} \cos^{- 1} x \\ = 2 (- \frac{1}{\sqrt{1 - x^{2}}}) \\ = - \frac{2}{\sqrt{1 - x^{2}}} \\ Thus, \\ \frac{d}{dx} \sec^{- 1} (\frac{1}{2 x^{2} - 1}) = - \frac{2}{\sqrt{1 - x^{2}}} . \end{array}$

Q.61

$Differentiate the following w . r . t . x : \frac{e^{x}}{sinx}$

Ans.

$\begin{array}{l} Let y = \frac{e^{x}}{sinx} \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \frac{e^{x}}{sinx} \\ = \frac{sinx \frac{d}{dx} e^{x} - e^{x} \frac{d}{dx} sinx}{{(sinx)}^{2}} [By Quotient Rule] \\ = \frac{sinx . e^{x} - e^{x} cosx}{{(sinx)}^{2}} \\ = \frac{e^{x} (sinx - cosx)}{\sin^{2} x}, x \neq nπ, n \in Z \end{array}$

Q.62

$D i f f e r e n t i a t e t h e f o l l o w i n g w . r . t . x : e^{s i n^{- 1} x}$

Ans.

$\begin{array}{l} Let y = e^{\sin^{- 1} x} \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} e^{\sin^{- 1} x} \\ = e^{\sin^{- 1} x} \frac{d}{dx} \sin^{- 1} x [By Chain Rule] \\ = e^{\sin^{- 1} x} \times \frac{1}{\sqrt{1 - x^{2}}} \\ = \frac{e^{\sin^{- 1} x}}{\sqrt{1 - x^{2}}}, x \in (- 1, 1) \end{array}$

Q.63

$D i f f e r e n t i a t e t h e f o l l o w i n g w . r . t . x : e^{x^{3}}$

Ans.

$\begin{array}{l} Let y = e^{x^{3}} \\ Differentiating both sides w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} e^{x^{3}} \\ = e^{x^{3}} \frac{d}{d x} x^{3} [By Chain Rule] \\ = e^{x^{3}} \times 3 x^{2} \\ = 3 x^{2} {.e}^{x^{3}} \end{array}$

Q.64

$Differentiate the following w . r . t . x : sin {(tan}^{- 1} e^{- x})$

Ans.

$\begin{array}{l} Let y = \sin (\tan^{- 1} e^{- x}) \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \sin (\tan^{- 1} e^{- x}) \\ = \cos (\tan^{- 1} e^{- x}) \frac{d}{dx} \tan^{- 1} e^{- x} [By Chain Rule] \\ = \cos (\tan^{- 1} e^{- x}) \times \frac{1}{1 + {(e^{- x})}^{2}} \times \frac{d}{dx} e^{- x} \\ = \frac{\cos (\tan^{- 1} e^{- x})}{1 + {(e^{- x})}^{2}} \times e^{- x} \frac{d}{dx} (- x) \\ = \frac{e^{- x} \cos (\tan^{- 1} e^{- x})}{1 + {(e^{- x})}^{2}} \times (- 1) \\ = - \frac{e^{- x} \cos (\tan^{- 1} e^{- x})}{1 + e^{- 2 x}} \end{array}$

Q.65

$Differentiate the following w . r . t . x : log {(cos e}^{x})$

Ans.

$\begin{array}{l} Let y = \log ({cose}^{x}) \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \log ({cose}^{x}) \\ = \frac{1}{{cose}^{x}} \times \frac{d}{dx} {cose}^{x} [By Chain Rule] \\ = \frac{1}{{cose}^{x}} \times - {sine}^{x} \times \frac{d}{dx} e^{x} \\ = \frac{- {sine}^{x}}{{cose}^{x}} \times e^{x} \\ = \frac{- e^{x} {sine}^{x}}{{cose}^{x}} \\ = - e^{x} {tane}^{x}, e^{x} \neq (2 n + 1) \frac{π}{2}, n \in N \end{array}$

Q.66

$\begin{array}{l} Differentiate the following w . r . t . x : \\ e^{x} + e^{x^{2}} + . .. + e^{x^{5}} \end{array}$

Ans.

$\begin{array}{l} Let y = e^{x} + e^{x^{2}} + . .. + e^{x^{5}} \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (e^{x} + e^{x^{2}} + . .. + e^{x^{5}}) \\ = \frac{d}{dx} e^{x} + \frac{d}{dx} e^{x^{2}} + \frac{d}{dx} e^{x^{3}} + \frac{d}{dx} e^{x^{4}} + \frac{d}{dx} e^{x^{5}} \\ = e^{x} + e^{x^{2}} \frac{d}{dx} (x^{2}) + e^{x^{3}} \frac{d}{dx} (x^{3}) + e^{x^{4}} \frac{d}{dx} (x^{4}) + e^{x^{5}} \frac{d}{dx} (x^{5}) \\ [Using chain rule] \\ = e^{x} + 2 {xe}^{x^{2}} + 3 x^{2} e^{x^{3}} + 4 x^{3} e^{x^{4}} + 5 x^{4} e^{x^{5}} \end{array}$

Q.67

$\begin{array}{l} Differentiate the following w . r . t . x : \\ \sqrt{e^{\sqrt{x}}}, x > 0 \end{array}$

Ans.

$\begin{array}{l} Let y = \sqrt{e^{\sqrt{x}}}, x > 0 \\ then, \\ y^{2} = e^{\sqrt{x}} \\ Differentiating both sides w . r . t . x, we get \\ \frac{d}{dx} y^{2} = \frac{d}{dx} (e^{\sqrt{x}}) \\ 2 y \frac{dy}{dx} = e^{\sqrt{x}} \frac{d}{dx} (\sqrt{x}) [Using chain rule] \\ \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{2 y} . \frac{1}{2 \sqrt{x}} \\ = \frac{e^{\sqrt{x}}}{4 \sqrt{x} \sqrt{e^{\sqrt{x}}}} \\ ∴ \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4 \sqrt{{xe}^{\sqrt{x}}}}, x > 0 \end{array}$

Q.68

$\begin{array}{l} Differentiate the following w . r . t . x : \\ \log (logx), x > 1 \end{array}$

Ans.

$\begin{array}{l} Let y = \log (logx), x > 1 \\ Differentiating g both sides w . r . t . x, we get \\ \frac{d}{dx} y = \frac{d}{dx} {\log (logx)} \\ \frac{dy}{dx} = \frac{1}{logx} \frac{d}{dx} (logx) [Using chain rule] \\ \frac{dy}{dx} = \frac{1}{logx} . \frac{1}{x} \\ \frac{dy}{dx} = \frac{1}{xlogx}, x > 1 \end{array}$

Q.69

$\begin{array}{l} Differentiate the following w . r . t . x : \\ \frac{cosx}{logx}, x > 0 \end{array}$

Ans.

$\begin{array}{l} Let y = \frac{cosx}{logx}, x > 0 \\ Differentiating both sides w . r . t . x, we get \\ \frac{d}{dx} y = \frac{d}{dx} (\frac{cosx}{logx}) \\ \frac{dy}{dx} = \frac{logx \frac{d}{dx} cosx - cosx \frac{d}{dx} logx}{{(logx)}^{2}} [By Quotient Rule] \\ \frac{dy}{dx} = \frac{logx \times - sinx - cosx \times \frac{1}{x}}{{(logx)}^{2}} \\ \frac{dy}{dx} = \frac{- (xlogx . sinx + cosx)}{x {(logx)}^{2}}, x > 0 \end{array}$

Q.70

$\begin{array}{l} D i f f e r e n t i a t e t h e f o l l o w i n g w . r . t . x : \\ c o s (l o g x + e^{x}), x > 0 \end{array}$

Ans.

$\begin{array}{l} L e t y = \cos (\log x + e^{x}), x > 0 \\ D i f f e r e n t i a t i n g both sides w .r .t . x, we get \\ \frac{d}{d x} y = \frac{d}{d x} \cos (\log x + e^{x}) \\ \frac{d y}{d x} = - \sin (\log x + e^{x}) \frac{d}{d x} (\log x + e^{x}) [Using c h a i n rule] \\ \frac{d y}{d x} = - \sin (\log x + e^{x}) (\frac{1}{x} + e^{x}) \\ \frac{d y}{d x} = - (\frac{1}{x} + e^{x}) . \sin (\log x + e^{x}), x > 0 \end{array}$

Q.71 Differentiate the function given with respect to x.
cosx.cos2x.cos3x

Ans.

$\begin{array}{l} Let ‹ y = cosx . \cos 2 x . \cos 3 x \\ Taking logarithm on both the sides, we get \\ logy = \log (cosx . \cos 2 x . \cos 3 x) \\ = logcosx + logcos 2 x + logcos 3 x \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} logy = \frac{d}{dx} logcosx + \frac{d}{dx} logcos 2 x + \frac{d}{dx} logcos 3 x \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{cosx} \frac{d}{dx} cosx + \frac{1}{\cos 2 x} \frac{d}{dx} \cos 2 x + \frac{1}{\cos 3 x} \frac{d}{dx} \cos 3 x \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{cosx} \times - sinx + \frac{1}{\cos 2 x} \times - \sin 2 x \frac{d}{dx} 2 x \\ + \frac{1}{\cos 3 x} \times - \sin 3 x \times \frac{d}{dx} 3 x \\ \frac{1}{y} \frac{dy}{dx} = - \frac{sinx}{cosx} - \frac{\sin 2 x}{\cos 2 x} \times 2 - \frac{\sin 3 x}{\cos 3 x} \times 3 \\ \frac{dy}{dx} = - y (tanx + 2 \tan 2 x + 3 \tan 3 x) \\ \frac{dy}{dx} = - cosx . \cos 2 x . \cos 3 x (tanx + 2 \tan 2 x + 3 \tan 3 x) \end{array}$

Q.72 Differentiate the function given with respect to x.

$\sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}}$

Ans.

$\begin{array}{l} Let y = \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} \\ Taking logarithm on both the sides, we get \\ logy = \log \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} \\ \Rightarrow logy = \frac{1}{2} \log {\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} \\ \Rightarrow logy = \frac{1}{2} {\log (x - 1) + \log (x - 2) - \log (x - 3) - \log (x - 4) - \log (x - 5)} \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} logy = \frac{1}{2} \frac{d}{dx} {\log (x - 1) + \log (x - 2) - \log (x - 3) - \log (x - 4) - \log (x - 5)} \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} (\begin{array}{l} \frac{1}{x - 1} \frac{d}{dx} (x - 1) + \frac{1}{x - 2} \frac{d}{dx} (x - 2) - \frac{1}{x - 3} \frac{d}{dx} (x - 3) \\ - \frac{1}{x - 4} \frac{d}{dx} (x - 4) - \frac{1}{x - 5} \frac{d}{dx} (x - 5) \end{array}) \\ [By chain rule] \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} (\frac{1}{x - 1} \times 1 + \frac{1}{x - 2} \times 1 - \frac{1}{x - 3} \times 1 - \frac{1}{x - 4} \times 1 - \frac{1}{x - 5} \times 1) \\ \frac{dy}{dx} = \frac{1}{2} y (\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}) \\ \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} (\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}) \end{array}$

Q.73 Differentiate the function given with respect to x.

${(logx)}^{cosx}$

Ans.

$\begin{array}{l} Let ‹ y = {(logx)}^{cosx} \\ Taking logarithm on both the sides, we get \\ logy = cosx . \log (logx) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} logy = \frac{d}{dx} {cosx . \log (logx)} \\ \frac{1}{y} \frac{dy}{dx} = cosx \frac{d}{dx} \log (logx) + \log (logx) \frac{d}{dx} cosx [By product rule] \\ \frac{1}{y} \frac{dy}{dx} = cosx . \frac{1}{logx} \frac{d}{dx} logx + \log (logx) . (- sinx) [By Chain rule] \\ \frac{dy}{dx} = y {\frac{cosx}{logx} \times \frac{1}{x} - sinx . \log (logx)} \\ ∴ \frac{dy}{dx} = {(logx)}^{cosx} {\frac{cosx}{xlogx} - sinx . \log (logx)} \end{array}$

Q.74 Differentiate the function given with respect to x.

$x^{x} - 2^{sinx}$

Ans.

$\begin{array}{l} Let ‹ y = x^{x} - 2^{sinx} \\ = y_{1} + y_{2} \\ So, y_{1} = x^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = \log (x^{x}) \\ = xlogx \\ Differentiating both sides with respect to x, we get \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} logx + logx \frac{d}{dx} x \\ = x . \frac{1}{x} + logx . 1 \\ = 1 + logx \\ \frac{{dy}_{1}}{dx} = y_{1} (1 + logx) \\ = x^{x} (1 + logx) \\ Now, y_{2} = 2^{sinx} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = \log (2^{sinx}) \\ = sinx . \log 2 \\ Differentiating both sides with respect to x, we get \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = \log 2 \frac{d}{dx} sinx \\ = \log 2 . cosx \\ \frac{{dy}_{1}}{dx} = y_{2} (\log 2 . cosx) \\ = 2^{sinx} cosx . \log 2 \\ Thus, \frac{dy}{dx} = \frac{{dy}_{1}}{dx} - \frac{{dy}_{2}}{dx} \\ = x^{x} (1 + logx) - 2^{sinx} cosx . \log 2 \end{array}$

Q.75 Differentiate the function given with respect to x.

${(x+3)}^{2} . {(x + 4)}^{3} . {(x + 5)}^{4}$

Ans.

$\begin{array}{l} Let ‹ y = {(x +3)}^{2} . {(x +4)}^{3} . {(x +5)}^{4} \\ Taking logarithm on both the sides, we get \\ logy = \log {{(x +3)}^{2} . {(x +4)}^{3} . {(x +5)}^{4}} \\ = 2 \log (x + 3) + 3 \log (x + 4) + 4 \log (x + 5) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} logy = 2 \frac{d}{dx} \log (x + 3) + 3 \frac{d}{dx} \log (x + 4) + 4 \frac{d}{dx} \log (x + 5) \\ \frac{1}{y} \frac{dy}{dx} = 2 \times \frac{1}{x + 3} \frac{d}{dx} (x + 3) + 3 \times \frac{1}{x + 4} \frac{d}{dx} (x + 4) + 4 \times \frac{1}{x + 5} \frac{d}{dx} (x + 5) \\ = \frac{2}{x + 3} \times 1 + \frac{3}{x + 4} \times 1 + \frac{4}{x + 5} \times 1 \\ \frac{dy}{dx} = y (\frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}) \\ = {(x +3)}^{2} . {(x +4)}^{3} . {(x +5)}^{4} (\frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}) \\ = {(x +3)}^{2} . {(x +4)}^{3} . {(x +5)}^{4} (\frac{2 (x + 4) (x + 5) + 3 (x + 3) (x + 5) + 4 (x + 3) (x + 4)}{(x + 3) (x + 4) (x + 5)}) \\ = (x +3) . {(x +4)}^{2} . {(x +5)}^{3} {\begin{cases} 2 (x^{2} + 9 x + 20) + 3 (x^{2} + 8 x + 15) \\ + 4 (x^{2} + 7 x + 12) \end{cases}} \\ = (x +3) . {(x +4)}^{2} . {(x +5)}^{3} {\begin{cases} 2 x^{2} + 18 x + 40 + 3 x^{2} + 24 x \\ + 45 + 4 x^{2} + 28 x + 48 \end{cases}} \\ ∴ \frac{dy}{dx} = (x +3) . {(x +4)}^{2} . {(x +5)}^{3} (9 x^{2} + 70 x + 133) \end{array}$

Q.76 Differentiate the function given with respect to x.

${(x+ \frac{1}{x})}^{x} + x^{(1 + \frac{1}{x})}$

Ans.

$\begin{array}{l} Let ‹ y = {(x + \frac{1}{x})}^{x} + x^{(1 + \frac{1}{x})} \\ = y_{1} + y_{2} \\ So, y_{1} = {(x + \frac{1}{x})}^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = xlog (x + \frac{1}{x}) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {xlog (x + \frac{1}{x})} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} \log (x + \frac{1}{x}) + \log (x + \frac{1}{x}) \frac{d}{dx} x \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x . \frac{1}{(x + \frac{1}{x})} \frac{d}{dx} (x + \frac{1}{x}) + \log (x + \frac{1}{x}) \times 1 \\ \frac{{dy}_{1}}{dx} = y_{1} {\frac{x^{2}}{x^{2} + 1} \times (1 - \frac{1}{x^{2}}) + \log (x + \frac{1}{x})} \\ \frac{{dy}_{1}}{dx} = {(x + \frac{1}{x})}^{x} {\frac{x^{2}}{x^{2} + 1} \times (\frac{x^{2} - 1}{x^{2}}) + \log (x + \frac{1}{x})} \\ \frac{{dy}_{1}}{dx} = {(x + \frac{1}{x})}^{x} {\frac{x^{2} - 1}{x^{2} + 1} + \log (x + \frac{1}{x})} \\ Now, y_{2} = x^{(1 + \frac{1}{x})} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = (1 + \frac{1}{x}) logx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{2} = \frac{d}{dx} {(1 + \frac{1}{x}) logx} \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = (1 + \frac{1}{x}) \frac{d}{dx} logx + logx \frac{d}{dx} (1 + \frac{1}{x}) [By ‹ product rule] \\ \frac{{dy}_{2}}{dx} = y_{2} {(1 + \frac{1}{x}) \frac{1}{x} + logx . (0 - \frac{1}{x^{2}})} \\ \frac{{dy}_{2}}{dx} = x^{(1 + \frac{1}{x})} {\frac{1}{x} + \frac{1}{x^{2}} - \frac{1}{x^{2}} logx} \\ = x^{(1 + \frac{1}{x})} {\frac{x + 1 - logx}{x^{2}}} \\ Thus, \\ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = {(x + \frac{1}{x})}^{x} {\frac{x^{2} - 1}{x^{2} + 1} + \log (x + \frac{1}{x})} + x^{(1 + \frac{1}{x})} {\frac{x + 1 - logx}{x^{2}}} \end{array}$

Q.77 Differentiate the function given with respect to x.

${(logx)}^{x} + x^{logx}$

Ans.

$\begin{array}{l} Let ‹ y = {(logx)}^{x} + x^{logx} \\ = y_{1} + y_{2} \\ So, y_{1} = {(logx)}^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = x . \log (logx) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {x . \log (logx)} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} \log (logx) + \log (logx) \frac{d}{dx} x [By Product Rule] \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x . \frac{1}{logx} \frac{d}{dx} logx + \log (logx) \times 1 \\ \frac{{dy}_{1}}{dx} = y_{1} {x . \frac{1}{logx} \frac{1}{x} + \log (logx)} \\ \frac{{dy}_{1}}{dx} = {(logx)}^{x} {\frac{1}{logx} + \log (logx)} \\ \frac{{dy}_{1}}{dx} = {(logx)}^{x -1} {1 + logx . \log (logx)} \\ Now, y_{2} = x^{logx} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = logx . logx \\ = {(logx)}^{2} \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{2} = \frac{d}{dx} {(logx)}^{2} \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = 2 logx \frac{d}{dx} logx [By Chain Rule] \\ = 2 logx \times \frac{1}{x} \\ \frac{{dy}_{2}}{dx} = \frac{2 y_{2}}{x} logx \\ = \frac{2 x^{logx}}{x} logx \\ = 2 x^{logx -1} logx \\ Hence, \\ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = {(logx)}^{x -1} {1 + logx . \log (logx)} + 2 x^{logx -1} logx \end{array}$

Q.78 Differentiate the function given with respect to x.

${(sinx)}^{x} + \sin^{- 1} \sqrt{x}$

Ans.

$\begin{array}{l} Let ‹ y = {(sinx)}^{x} + \sin^{- 1} \sqrt{x} \\ = y_{1} + y_{2} \\ So, y_{1} = {(sinx)}^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = x . \log (sinx) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {x . \log (sinx)} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} \log (sinx) + \log (sinx) \frac{d}{dx} x [By Product Rule] \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x . \frac{1}{sinx} \frac{d}{dx} sinx + \log (sinx) \times 1 \\ \frac{{dy}_{1}}{dx} = y_{1} {x . \frac{1}{sinx} cosx + \log (sinx)} \\ \frac{{dy}_{1}}{dx} = {(sinx)}^{x} {x \frac{cosx}{sinx} + \log (sinx)} \\ \frac{{dy}_{1}}{dx} = {(sinx)}^{x} {xcotx + \log (sinx)} \\ Now, y_{2} = \sin^{- 1} \sqrt{x} \\ Differentiating both sides with respect to x, we get \\ \frac{{dy}_{2}}{dx} = \frac{d}{dx} \sin^{- 1} \sqrt{x} \\ \frac{{dy}_{2}}{dx} = \frac{1}{\sqrt{1 - {(\sqrt{x})}^{2}}} \frac{d}{dx} \sqrt{x} [By Chain Rule] \\ = \frac{1}{\sqrt{1 - x}} \times \frac{1}{2 \sqrt{x}} \\ \frac{{dy}_{2}}{dx} = \frac{1}{2 \sqrt{x - x^{2}}} \\ Hence, \\ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = {(sinx)}^{x} {xcotx + \log (sinx)} + \frac{1}{2 \sqrt{x - x^{2}}} \end{array}$

Q.79 Differentiate the function given with respect to x.

$x^{sinx} + {(sinx)}^{cosx}$

Ans.

$\begin{array}{l} Let ‹ y = x^{sinx} + {(sinx)}^{cosx} \\ = y_{1} + y_{2} \\ So, y_{1} = x^{sinx} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = sinx . logx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {sinx . logx} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = sinx \frac{d}{dx} logx + logx \frac{d}{dx} sinx [By Product Rule] \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = sinx . \frac{1}{x} + logx \times cosx \\ \frac{{dy}_{1}}{dx} = y_{1} {sinx . \frac{1}{x} + logx \times cosx} \\ \frac{{dy}_{1}}{dx} = x^{sinx} {\frac{sinx}{x} + cosx . logx} \\ Now, y_{2} = {(sinx)}^{cosx} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = cosx . logsinx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} (cosx . logsinx) \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = cosx \frac{d}{dx} logsinx + logsinx \frac{d}{dx} cosx [By Product Rule] \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = cosx . \frac{1}{sinx} \frac{d}{dx} sinx + logsinx \times - sinx \\ \frac{{dy}_{2}}{dx} = {(sinx)}^{cosx} (cosx . \frac{1}{sinx} . cosx - sinx . logsinx) \\ = {(sinx)}^{cosx} (cotx . cosx - sinx . logsinx) \\ ∴ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = x^{sinx} (\frac{sinx}{x} + cosx . logx) + {(sinx)}^{cosx} (cosx . cotx - sinx . logsinx) \end{array}$

Q.80 Differentiate the function given with respect to x.

$x^{xcosx} + \frac{x^{2} + 1}{x^{2} - 1}$

Ans.

$\begin{array}{l} Let y = x^{xcosx} + \frac{x^{2} + 1}{x^{2} - 1} \\ = y_{1} + y_{2} \\ So, y_{1} = x^{xcosx} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = xcosx . logx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {xcosx . logx} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = xcosx \frac{d}{dx} logx + logx \frac{d}{dx} (xcosx) \\ [By Product Rule] \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = xcosx . \frac{1}{x} + logx (x \frac{d}{dx} cosx + cosx \frac{d}{dx} x) \\ [By Product Rule] \\ \frac{{dy}_{1}}{dx} = y_{1} {cosx + logx (- xsinx + cosx)} \\ \frac{{dy}_{1}}{dx} = x^{xcosx} {cosx - xsinxlogx + logx . cosx} \\ \frac{{dy}_{1}}{dx} = x^{xcosx} {cosx (1 + logx) - xsinxlogx} \\ Now, \\ y_{2} = \frac{x^{2} + 1}{x^{2} - 1} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = \log (x^{2} + 1) - \log (x^{2} - 1) \\ Differentiating w . r . t . x, we get \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = \frac{1}{x^{2} + 1} \frac{d}{dx} (x^{2} + 1) - \frac{1}{x^{2} - 1} \frac{d}{dx} (x^{2} - 1) [By chain rule] \\ \frac{{dy}_{2}}{dx} = y_{2} (\frac{1}{x^{2} + 1} \times 2 x - \frac{1}{x^{2} - 1} \times 2 x) \\ = 2 x (\frac{x^{2} + 1}{x^{2} - 1}) (\frac{1}{x^{2} + 1} - \frac{1}{x^{2} - 1}) \\ = 2 x (\frac{x^{2} + 1}{x^{2} - 1}) (\frac{(x^{2} - 1) - (x^{2} + 1)}{(x^{2} + 1) ((x^{2} - 1))}) \\ = 2 x (\frac{x^{2} + 1}{x^{2} - 1}) {\frac{x^{2} - 1 - x^{2} - 1}{(x^{2} + 1) (x^{2} - 1)}} \\ = 2 x (\frac{1}{x^{2} - 1}) {\frac{- 2}{(x^{2} - 1)}} \\ \frac{{dy}_{2}}{dx} = \frac{- 4 x}{{(x^{2} - 1)}^{2}} \\ Thus, \\ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = x^{xcosx} {cosx (1 + logx) - xsinxlogx} - \frac{4 x}{{(x^{2} - 1)}^{2}} \end{array}$

Q.81

$Differentiate the function given with respect to x. {(xcosx)}^{x} + {(xsinx)}^{\frac{1}{x}}$

Ans.

$\begin{array}{l} Let ‹ y = {(xcosx)}^{x} + {(xsinx)}^{\frac{1}{x}} \\ = y_{1} + y_{2} \\ So, y_{1} = {(xcosx)}^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = xlog (xcosx) \\ = xlogx + xlogcosx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} (x . logx + x . logcosx) \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = \frac{d}{dx} (x . logx) + \frac{d}{dx} (x . logcosx) \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} logx + logx \frac{d}{dx} x + x \frac{d}{dx} logcosx + logcosx \frac{d}{dx} x \\ [By Product Rule] \\ \frac{{dy}_{1}}{dx} = y_{1} (x . \frac{1}{x} + logx + x \frac{1}{cosx} . \frac{d}{dx} cosx + logcosx) \\ \frac{{dy}_{1}}{dx} = {(xcosx)}^{x} (1 + logx + x \frac{1}{cosx} \times - sinx + logcosx) \\ = {(xcosx)}^{x} (1 + logx - x \frac{sinx}{cosx} + logcosx) \\ \frac{{dy}_{1}}{dx} = {(xcosx)}^{x} (1 + logx - xtanx + logcosx) \\ Now, \\ y_{2} = {(xsinx)}^{\frac{1}{x}} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = \frac{1}{x} \log (xsinx) \\ Differentiating w . r . t . x, we get \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = \frac{1}{x} \frac{d}{dx} \log (xsinx) + \log (xsinx) \frac{d}{dx} . \frac{1}{x} [By product rule] \\ \frac{{dy}_{2}}{dx} = y_{2} {\frac{1}{x} (\frac{1}{xsinx}) \frac{d}{dx} (xsinx) - \log (xsinx) . \frac{1}{x^{2}}} [By chain rule] \\ \frac{{dy}_{2}}{dx} = {(xsinx)}^{\frac{1}{x}} {\frac{1}{x} (\frac{1}{xsinx}) (x \frac{d}{dx} sinx + sinx \frac{d}{dx} x) - \log (xsinx) . \frac{1}{x^{2}}} \\ \frac{{dy}_{2}}{dx} = {(xsinx)}^{\frac{1}{x}} {\frac{1}{x} (\frac{1}{xsinx}) (x . cosx + sinx) - \log (xsinx) . \frac{1}{x^{2}}} \\ \frac{{dy}_{2}}{dx} = {(xsinx)}^{\frac{1}{x}} {(\frac{x . cosx + sinx}{x^{2} sinx}) - \log (xsinx) . \frac{1}{x^{2}}} \\ ∴ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ \frac{dy}{dx} = {(xcosx)}^{x} (1 + logx - xtanx + logcosx) \\ + {(xsinx)}^{\frac{1}{x}} {(\frac{x . cosx + sinx}{x^{2} sinx}) - \log (xsinx) . \frac{1}{x^{2}}} \end{array}$

Q.82

$Find \frac{dy}{dx} of the function x^{y} + y^{x} =$

Ans.

$\begin{array}{l} The given function is \\ x^{y} + y^{x} = 1 \\ Let u = x^{y} and v = y^{x} \\ Then, function becomes \\ u + v = 1 \\ Differentiating w . r . t . x, we get \\ \frac{du}{dx} + \frac{dv}{dx} = 0 . .. (i) \\ Since, u = x^{y} \\ Taking \log both sides, we get \\ logu = {logx}^{y} \\ logu = ylogx \\ Differentiating both sides w . r . t . x, we get \\ \frac{d}{dx} logu = \frac{d}{dx} ylogx \\ \frac{1}{u} \frac{du}{dx} = y \frac{d}{dx} logx + logx \frac{dy}{dx} \\ = y \times \frac{1}{x} + logx \frac{dy}{dx} \\ \frac{du}{dx} = u (\frac{y}{x} + logx \frac{dy}{dx}) \\ \frac{du}{dx} = x^{y} (\frac{y}{x} + logx \frac{dy}{dx}) . .. (ii) \\ And v = y^{x} \\ Taking \log both sides, we get \\ logv = {logy}^{x} \\ logv = xlogy \\ Differentiating both sides w . r . t . x, we get \\ \frac{d}{dx} logv = \frac{d}{dx} xlogy \\ \frac{1}{v} \frac{dv}{dx} = x \frac{d}{dx} logy + logy \frac{d}{dx} x \\ = x \times \frac{1}{y} \frac{dy}{dx} + logy (1) \\ \frac{dv}{dx} = v (\frac{x}{y} \frac{dy}{dx} + logy) \\ \frac{dv}{dx} = y^{x} (\frac{x}{y} \frac{dy}{dx} + logy) . .. (iii) \\ From equatin (i), (ii) and (iii), we have \\ x^{y} (\frac{y}{x} + logx \frac{dy}{dx}) + y^{x} (\frac{x}{y} \frac{dy}{dx} + logy) = 0 \\ x^{y - 1} y + x^{y} logx \frac{dy}{dx} + {yx}^{- 1} x \frac{dy}{dx} + y^{x} logy = 0 \\ (x^{y} logx + {yx}^{- 1} x) \frac{dy}{dx} = - x^{y - 1} y - y^{x} logy \\ \frac{dy}{dx} = - \frac{{yx}^{y - 1} + y^{x} logy}{x^{y} logx + {xy}^{x}^{- 1}} \end{array}$

Q.83

$Find \frac{dy}{dx} of the function : y^{x} = x^{y}$

Ans.

$\begin{array}{l} The given function is y^{x} = x^{y} \\ Taking logarithm both sides, we get \\ {logy}^{x} = {logx}^{y} \\ xlogy = ylogx \\ Differentiating both sides, w . r . t . x, we get \\ x \frac{d}{dx} logy + logy \frac{d}{dx} x = y \frac{d}{dx} logx + logx \frac{d}{dx} y [By Product Rule] \\ \frac{x}{y} \frac{dy}{dx} + logy \times 1 = \frac{y}{x} + logx \frac{dy}{dx} \\ \frac{x}{y} \frac{dy}{dx} - logx \frac{dy}{dx} = \frac{y}{x} - logy \\ (\frac{x}{y} - logx) \frac{dy}{dx} = \frac{y}{x} - logy \\ \frac{dy}{dx} = \frac{(\frac{y}{x} - logy)}{(\frac{x}{y} - logx)} \\ = \frac{y (y - xlogy)}{x (x - ylogx)} \end{array}$

Q.84

$Find \frac{dy}{dx} of the function : {(cosx)}^{y} = {(cosy)}^{x}$

Ans.

$\begin{array}{l} The given function is {(cosx)}^{y} = {(cosy)}^{x} \\ Taking logarithm both sides, we get \\ \log {(cosx)}^{y} = \log {(cosy)}^{x} \\ ylogcosx = xlogcosy \\ Differentiating both sides, w . r . t . x, we get \\ y \frac{d}{dx} logcosx + logcosx \frac{d}{dx} y = x \frac{d}{dx} logcosy + logcosy \frac{d}{dx} x [By Product Rule] \\ y \times \frac{1}{cosx} \frac{d}{dx} cosx + logcosx \frac{dy}{dx} = x \times \frac{1}{cosy} \frac{d}{dx} cosy + logcosy \\ \frac{y}{cosx} \times - sinx + logcosx \frac{dy}{dx} = \frac{x}{cosy} \times - siny \frac{dy}{dx} + logcosy \\ \frac{- ysinx}{cosx} + logcosx \frac{dy}{dx} = \frac{- xsiny}{cosy} \frac{dy}{dx} + logcosy \\ logcosx \frac{dy}{dx} + \frac{xsiny}{cosy} \frac{dy}{dx} = logcosy + \frac{ysinx}{cosx} \\ (logcosx + \frac{xsiny}{cosy}) \frac{dy}{dx} = logcosy + \frac{ysinx}{cosx} \\ \frac{dy}{dx} = \frac{(logcosy + \frac{ysinx}{cosx})}{(logcosx + \frac{xsiny}{cosy})} \\ = \frac{(logcosy + ytanx)}{(logcosx + xtany)} \end{array}$

Q.85

$Find \frac{dy}{dx} of the function : xy = e^{(x - y)}$

Ans.

$\begin{array}{l} The given function is xy = e^{(x - y)} \\ Taking logarithm both sides, we get \\ logxy = {loge}^{(x - y)} \\ \Rightarrow logx + logy = (x - y) loge \\ \Rightarrow logx + logy = (x - y) \times 1 âˆµ [loge = 1] \\ \Rightarrow logx + logy = (x - y) \\ Differentiating both sides, w . r . t . x, we get \\ \frac{d}{dx} logx + \frac{d}{dx} logy = \frac{d}{dx} x - \frac{dy}{dx} [By Product Rule] \\ \Rightarrow \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx} \\ \Rightarrow (1 + \frac{1}{y}) \frac{dy}{dx} = 1 - \frac{1}{x} \\ \Rightarrow \frac{dy}{dx} = \frac{1 - \frac{1}{x}}{(1 + \frac{1}{y})} \\ = \frac{(\frac{x - 1}{x})}{(\frac{y + 1}{y})} \\ \Rightarrow \frac{dy}{dx} = \frac{y (x - 1)}{x (y + 1)} \end{array}$

Q.86

$\begin{array}{l} Find the derivative of the function given by \\ f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) \\ and hence find f ‘ (1) . \end{array}$

Ans.

$\begin{array}{l} The given relationship is f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) \\ Taking logarithm on both the sides, we get \\ logf (x) = \log (1 + x) + \log (1 + x^{2}) + \log (1 + x^{4}) + \log (1 + x^{8}) \\ Differentiating both sides with respect to x, we get \\ \frac{1}{f (x)} \frac{d}{dx} f (x) = \frac{1}{1 + x} \frac{d}{dx} (1 + x) + \frac{1}{1 + x^{2}} \frac{d}{dx} (1 + x^{2}) + \frac{1}{1 + x^{4}} \frac{d}{dx} (1 + x^{4}) \\ + \frac{1}{1 + x^{8}} \frac{d}{dx} (1 + x^{8}) \\ = \frac{1}{1 + x} (1) + \frac{1}{1 + x^{2}} (0 + 2 x) + \frac{1}{1 + x^{4}} (0 + 4 x^{3}) \\ + \frac{1}{1 + x^{8}} (0 + 8 x^{7}) \\ \frac{d}{dx} f (x) = f (x) (\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}) \\ ∴ f ‘ (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}) \\ Putting x = 1, we get \\ f ‘ (1) = (1 + 1) (1 + 1^{2}) (1 + 1^{4}) (1 + 1^{8}) (\frac{1}{1 + 1} + \frac{2 (1)}{1 + 1^{2}} + \frac{4 {(1)}^{3}}{1 + 1^{4}} + \frac{8 {(1)}^{7}}{1 + 1^{8}}) \\ = 2 \times 2 \times 2 \times 2 (\frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2}) \\ = 16 (\frac{15}{2}) \\ ∴ f ‘ (1) = 120 \end{array}$

Q.87 Differentiate (x² – 5x + 8)(x³ + 7x + 9) in three ways mentioned below:

(i) By using product rule.

(ii) By expanding the product to obtain a single polynomial.

(iii) By logarithmic differentiation. Do they all give the same answer?

Ans.

$\begin{array}{l} Let y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) \\ (i) Differentiating w . r . t . x, by using product rule, we get \\ \frac{dy}{dx} = \frac{d}{dx} {(x^{2} - 5 x + 8) (x^{3} + 7 x + 9)} \\ = (x^{2} - 5 x + 8) \frac{d}{dx} (x^{3} + 7 x + 9) + (x^{3} + 7 x + 9) \frac{d}{dx} (x^{2} - 5 x + 8) \\ = (x^{2} - 5 x + 8) (3 x^{2} + 7) + (x^{3} + 7 x + 9) (2 x - 5) \\ = 3 x^{4} + 7 x^{2} - 15 x^{3} - 35 x + 24 x^{2} + 56 + 2 x^{4} - 5 x^{3} + 14 x^{2} - 35 x + 18 x - 45 \\ ∴ \frac{dy}{dx} = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11 \\ (ii) y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) \\ = x^{5} + 7 x^{3} + 9 x^{2} - 5 x^{4} - 35 x^{2} - 45 x + 8 x^{3} + 56 x + 72 \\ = x^{5} - 5 x^{4} + 15 x^{3} - 26 x^{2} + 11 x + 72 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{5} - 5 x^{4} + 15 x^{3} - 26 x^{2} + 11 x + 72) \\ = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11 \\ (iii) y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) \\ Taking logarithm on both the sides, we get \\ logy = \log {(x^{2} - 5 x + 8) (x^{3} + 7 x + 9)} \\ = \log (x^{2} - 5 x + 8) + \log (x^{3} + 7 x + 9) \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} logy = \frac{d}{dx} \log (x^{2} - 5 x + 8) + \frac{d}{dx} \log (x^{3} + 7 x + 9) \\ = \frac{1}{(x^{2} - 5 x + 8)} \frac{d}{dx} (x^{2} - 5 x + 8) + \frac{1}{(x^{3} + 7 x + 9)} \frac{d}{dx} (x^{3} + 7 x + 9) \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{(x^{2} - 5 x + 8)} (2 x - 5) + \frac{1}{(x^{3} + 7 x + 9)} (3 x^{2} + 7) \\ \frac{dy}{dx} = y (\frac{(2 x - 5) (x^{3} + 7 x + 9) + (3 x^{2} + 7) (x^{2} - 5 x + 8)}{(x^{2} - 5 x + 8) (x^{3} + 7 x + 9)}) \\ \frac{dy}{dx} = y (\frac{(2 x - 5) (x^{3} + 7 x + 9) + (3 x^{2} + 7) (x^{2} - 5 x + 8)}{y}) \\ [âˆµ Let y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) \\ \frac{dy}{dx} = 2 x^{4} + 14 x^{2} + 18 x - 5 x^{3} - 35 x - 45 + 3 x^{4} - 15 x^{3} \\ + 24 x^{2} + 7 x^{2} - 35 x + 56 \\ \frac{dy}{dx} = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11 \\ From the above three observations, it can be say that all the \\ results are same . \end{array}$

Q.88

$\begin{array}{l} If u, v and w are functions of x, ‹ then show that \\ \frac{d}{dx} (u, v, w) = \frac{du}{dx} v . w + u \frac{dv}{dx} . w + u . v . \frac{dw}{dx} \\ in two ways - first by repeated application of product rule, \\ second by logarithmic differentiation . \end{array}$

Ans.

$\begin{array}{l} Let y = u . v . w = u . (v . w) \\ By applying product rule, we get \\ \frac{dy}{dx} = \frac{d}{dx} {u . (v . w)} \\ = u \frac{d}{dx} (v . w) + (v . w) \frac{d}{dx} u \\ = u {v \frac{d}{dx} w + w \frac{d}{dx} v} + (v . w) \frac{d}{dx} u \\ = u . v \frac{d}{dx} w + u . w \frac{d}{dx} v + (v . w) \frac{d}{dx} u \\ \frac{dy}{dx} = \frac{du}{dx} . v . w + \frac{dv}{dx} . u . w + \frac{dw}{dx} . u . v \\ Thus, this is the differentiation by product rule . \\ By taking logarithm on both sides of the equation y = u . v . w, we get \\ logy = logu + logv + logw \\ Differentiating both sides with respect to x, we get \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \\ \frac{dy}{dx} = y (\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}) \\ = u . v . w (\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}) \\ \frac{dy}{dx} = \frac{du}{dx} . v . w + \frac{dv}{dx} . u . w + \frac{dw}{dx} . u . v \\ Thus, this is the differentiation by logarithmic differentiation . \end{array}$

Q.89 If x and y are connected parametrically by the given equations, without eliminating the parameter, find dy/dx.
x = 2at², y = at⁴

Ans.

$\begin{array}{l} The given equations are x = 2 {at}^{2}, y = {at}^{4} \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} 2 {at}^{2} = 4 at \\ and \\ \frac{dy}{dt} = \frac{d}{dt} {at}^{4} = 4 {at}^{3} \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{4 {at}^{3}}{4 at} = t^{2} \end{array}$

Q.90

$\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = a cosθ, y = b cosθ \end{array}$

Ans.

$\begin{array}{l} The given equations are x = acosθ, y = bcosθ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} acosθ = - asinθ \\ and \\ \frac{dy}{dθ} = \frac{d}{dθ} bcosθ = - bsinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{- bsinθ}{- asinθ} = \frac{b}{a} \end{array}$

Q.91

$\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = sint, y = \cos 2 t \end{array}$

Ans.

$\begin{array}{l} The given equations are x = sint, y = \cos 2 t \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} sint = cost \\ and \\ \frac{dy}{dt} = \frac{d}{dt} \cos 2 t \\ = - 2 \sin 2 t \\ = - 4 sintcost \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{- 4 sintcost}{cost} \\ = - 4 sint \end{array}$

Q.92

$\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x= 4 t, y = \frac{4}{t} \end{array}$

Ans.

$\begin{array}{l} The given equations are x = 4 t, y = \frac{4}{t} \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} 4 t = 4 \\ and \\ \frac{dy}{dt} = \frac{d}{dt} \frac{4}{t} = - \frac{4}{t^{2}} \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{- \frac{4}{t^{2}}}{4} \\ ∴ \frac{dy}{dx} = - \frac{1}{t^{2}} \end{array}$

Q.93

$\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = cosθ - \cos 2 θ y = sinθ - \sin 2 θ \end{array}$

Ans.

$\begin{array}{l} The given equations are x = \cos θ - \cos 2 θ, y = \sin θ - \sin 2 θ \\ Differentiating w .r .t . θ, we get \\ \frac{d x}{d θ} = \frac{d}{d θ} (\cos θ - \cos 2 θ) \\ = - \sin θ + 2 \sin 2 θ \\ a n d \\ \frac{d y}{d θ} = \frac{d}{d θ} (\sin θ - \sin 2 θ) \\ = \cos θ - 2 \cos 2 θ \\ ∴ \frac{d y}{d x} = \frac{(\frac{d y}{d θ})}{(\frac{d x}{d θ})} \\ = \frac{\cos θ - 2 \cos 2 θ}{2 \sin 2 θ - \sin θ} \end{array}$

Q.94

$\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = a (θ - sinθ), y = a (1 + cosθ) \end{array}$

Ans.

$\begin{array}{l} The given equations are x = a (θ - sinθ), y = a (1 + cosθ) \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} a (θ - sinθ) \\ = a (1 - cosθ) \\ and \\ \frac{dy}{dθ} = \frac{d}{dθ} a (1 + cosθ) \\ = a (0 - sinθ) \\ = - asinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{- asinθ}{a (1 - cosθ)} \\ = - \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} \\ = - \cot \frac{θ}{2} \end{array}$

Q.95

$\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}} y = \frac{\cos^{3} t}{\sqrt{\cos 2 t}} \end{array}$

Ans.

$\begin{array}{l} The given equations are x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}}, y = \frac{\cos^{3} t}{\sqrt{\cos 2 t}} \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} \frac{\sin^{3} t}{\sqrt{\cos 2 t}} \\ = \frac{\sqrt{\cos 2 t} \frac{d}{dt} \sin^{3} t - \sin^{3} t \frac{d}{dt} {(\cos 2 t)}^{\frac{1}{2}}}{{(\sqrt{\cos 2 t})}^{2}} \\ = \frac{\sqrt{\cos 2 t} (3 \sin^{2} t \frac{d}{dt} sint) - \sin^{3} t \times \frac{1}{2} {(\cos 2 t)}^{- \frac{1}{2}} \frac{d}{dt} \cos 2 t}{\cos 2 t} \\ = \frac{\sqrt{\cos 2 t} (3 \sin^{2} t \times cost) - \sin^{3} t \times \frac{1}{2} {(\cos 2 t)}^{- \frac{1}{2}} \times - 2 \sin 2 t}{\cos 2 t} \\ = \frac{3 \cos 2 {tsin}^{2} tcost + \sin^{3} tsin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \\ and \\ \frac{dy}{dt} = \frac{d}{dt} \frac{\cos^{3} t}{\sqrt{\cos 2 t}} \\ = \frac{\sqrt{\cos 2 t} \frac{d}{dt} \cos^{3} t - \cos^{3} t \frac{d}{dt} {(\cos 2 t)}^{\frac{1}{2}}}{{(\sqrt{\cos 2 t})}^{2}} \\ = \frac{\sqrt{\cos 2 t} (3 \cos^{2} t \frac{d}{dt} cost) - \cos^{3} t \times \frac{1}{2} {(\cos 2 t)}^{- \frac{1}{2}} \frac{d}{dt} \cos 2 t}{\cos 2 t} \\ = \frac{\sqrt{\cos 2 t} (3 \cos^{2} t \times - sint) - \cos^{3} t \times \frac{1}{2} {(\cos 2 t)}^{- \frac{1}{2}} \times - 2 \sin 2 t}{\cos 2 t} \\ = \frac{- 3 \cos 2 {tcos}^{2} tsint + \cos^{3} tsin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{\frac{- 3 \cos 2 {tcos}^{2} tsint + \cos^{3} tsin 2 t}{\cos 2 t \sqrt{\cos 2 t}}}{\frac{3 \cos 2 {tsin}^{2} tcost + \sin^{3} tsin 2 t}{\cos 2 t \sqrt{\cos 2 t}}} \\ ∴ \frac{dy}{dx} = \frac{- 3 \cos 2 {tcos}^{2} tsint + \cos^{3} t \times 2 sintcost}{3 \cos 2 {tsin}^{2} tcost + \sin^{3} t \times 2 sintcost} [âˆµ \sin 2 t = 2 sintcost] \\ = \frac{sintcost (- 3 \cos 2 tcost + 2 \cos^{3} t)}{sintcost (3 \cos 2 tsint + 2 \sin^{3} t)} \\ = \frac{- 3 (2 \cos^{2} t - 1) cost + 2 \cos^{3} t}{3 (1 - 2 \sin^{2} t) sint + 2 \sin^{3} t} \\ = \frac{- 6 \cos^{3} t + 3 cost + 2 \cos^{3} t}{3 sint - 6 \sin^{3} t + 2 \sin^{3} t} \\ = \frac{- 4 \cos^{3} t + 3 cost}{3 sint - 4 \sin^{3} t} \\ = - \frac{\cos 3 t}{\sin 3 t} \end{array}$

Q.96

$\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = a (\cos t + \log \tan \frac{t}{2}), y = a \sin t \end{array}$

Ans.

$\begin{array}{l} The given equations are x = a (cost + logtan \frac{t}{2}), y = asint \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} a (cost + logtan \frac{t}{2}) \\ = a (\frac{d}{dt} cost + \frac{d}{dt} logtan \frac{t}{2}) \\ = a (- sint + \frac{1}{\tan (\frac{t}{2})} \frac{d}{dt} \tan (\frac{t}{2})) [By chain rule] \\ = a (- sint + \frac{1}{\tan (\frac{t}{2})} . \sec^{2} (\frac{t}{2}) \frac{d}{dt} \frac{t}{2}) \\ = a (- sint + \frac{1}{\tan (\frac{t}{2})} . \sec^{2} (\frac{t}{2}) \times \frac{1}{2}) \\ = a (- sint + \frac{1}{2} \frac{1}{\sin (\frac{t}{2}) \cos (\frac{t}{2})}) \\ = a (- sint + \frac{1}{\sin 2 (\frac{t}{2})}) [âˆµ \sin 2 θ = 2 sinθcosθ] \\ = a (- sint + \frac{1}{sint}) \\ = a (\frac{1 - \sin^{2} t}{sint}) \\ = a \frac{\cos^{2} t}{sint} \\ and \\ \frac{dy}{dt} = \frac{d}{dt} asint \\ = acost \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{acost}{(a \frac{\cos^{2} t}{sint})} \\ = \frac{sint}{cost} \\ ∴ \frac{dy}{dx} = tant \end{array}$

Q.97 Ifxandyareconnectedparametricallybythebelowequations, withouteliminatingtheparameter,find dy dx.
x = asecθ,y = btanθ

Ans.

$\begin{array}{l} The given equations are x = asecθ, y = b tanθ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} asecθ \\ = asecθtanθ \\ and \\ \frac{dy}{dθ} = \frac{d}{dθ} b tanθ \\ = {bsec}^{2} θ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{{bsec}^{2} θ}{asecθtanθ} \\ = \frac{b}{a} . secθ . cotθ \\ = \frac{b}{a} . \frac{1}{cosθ} . \frac{cosθ}{sinθ} \\ = \frac{b}{a} \times \frac{1}{sinθ} \\ = \frac{b}{a} cosecθ \end{array}$

Q.98

$\begin{array}{l} If x and y are connected parametrically by the below equations, \\ without eliminating the parameter, find \frac{dy}{dx} . \\ x = a (cosθ + θ sinθ), y = a (sinθ - θ cosθ) \end{array}$

Ans.

$\begin{array}{l} The given equations are x = a (cosθ + θsinθ), y = a (sinθ - θcos θ) \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} a (cosθ + θsinθ) \\ = a (- sinθ + θ \frac{d}{dθ} sinθ + sinθ \frac{d}{dθ} θ) \\ = a (- sinθ + θcosθ + sinθ \times 1) \\ = aθcosθ \\ and \\ \frac{dy}{dθ} = \frac{d}{dθ} a (sinθ - θcos θ) \\ = a (cosθ - θ \frac{d}{dθ} cosθ - cosθ \frac{d}{dθ} θ) \\ = a (cosθ + θsinθ - cosθ \times 1) \\ = aθsinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} = \frac{aθsinθ}{aθcosθ} = tanθ \end{array}$

Q.99

$\sqrt{a^{\sin^{- 1} t}}, y = \sqrt{a^{\cos^{- 1} t}}, show that \frac{dy}{dx} = - \frac{y}{x}$

Ans.

$\begin{array}{l} The given equations are x = \sqrt{a^{\sin^{- 1} t}} = a^{\frac{1}{2} \sin^{- 1} t}, \\ y = \sqrt{a^{\cos^{- 1} t}} = a^{\frac{1}{2} \cos^{- 1} t} \\ Taking logarithm both sides, we get \\ logx = {loga}^{\frac{1}{2} \sin^{- 1} t} \\ = \frac{1}{2} \sin^{- 1} t . loga \\ and \\ logy = {loga}^{\frac{1}{2} \cos^{- 1} t} \\ = \frac{1}{2} \cos^{- 1} t . loga \\ Differentiating w . r . t . t, we get \\ \frac{d}{dt} logx = \frac{d}{dt} \frac{1}{2} \sin^{- 1} t . loga \\ \frac{1}{x} \frac{dx}{dt} = \frac{1}{2} loga \frac{d}{dt} \sin^{- 1} t \\ = \frac{1}{2} loga \times \frac{1}{\sqrt{1 - t^{2}}} \\ \frac{dx}{dt} = x (\frac{loga}{2 \sqrt{1 - t^{2}}}) \\ and \\ \frac{d}{dt} logy = \frac{d}{dt} \frac{1}{2} \cos^{- 1} t . loga \\ \frac{1}{y} \frac{dy}{dt} = \frac{1}{2} loga \frac{d}{dt} \cos^{- 1} t \\ = \frac{1}{2} loga \times \frac{- 1}{\sqrt{1 - t^{2}}} \\ \frac{dy}{dt} = - y (\frac{loga}{2 \sqrt{1 - t^{2}}}) \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{- y (\frac{loga}{2 \sqrt{1 - t^{2}}})}{x (\frac{loga}{2 \sqrt{1 - t^{2}}})} \\ \Rightarrow \frac{dy}{dx} = - \frac{y}{x} \\ Hence, \frac{dy}{dx} = - \frac{y}{x} \end{array}$

Q.100 Find the second order derivative of the function:
x² + 3x + 2

Ans.

$\begin{array}{l} Let y = x^{2} + 3 x + 2 \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{2} + 3 x + 2) \\ = 2 x + 3 \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (2 x + 3) \\ = 2 \\ Thus, second derivative of given function is 2 . \end{array}$

Q.101

${Findthesecondorderderivativeof‹thefunction:x}^{20}$

Ans.

$\begin{array}{l} Let y = x^{20} \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} x^{20} \\ = 20 x^{19} \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (20 x^{19}) \\ = 20 \times 19 x^{18} \\ = 380 x^{18} \\ Thus, second derivative of given function is 380 x^{18} . \end{array}$

Q.102

$\begin{array}{l} Find the second order derivative of the function : \\ x . cosx \end{array}$

Ans.

$\begin{array}{l} Let y = xcosx \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (xcosx) \\ = x \frac{d}{dx} cosx + cosx \frac{d}{dx} x [By product rule] \\ = x (- sinx) + cosx \times 1 \\ = - xsinx + cosx \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (- xsinx + cosx) \\ = - (x \frac{d}{dx} sinx + sinx \frac{d}{dx} x) + \frac{d}{dx} cosx \\ = - (xcosx + sinx \times 1) - sinx \\ = - xcosx - sinx - sinx \\ = - xcosx - 2 sinx \\ Thus, second derivative of given function is - (xcosx + 2 sinx) . \end{array}$

Q.103

$Findthesecondorderderivative‹of‹thefunction:logx$

Ans.

$\begin{array}{l} Let y = logx \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} logx \\ = \frac{1}{x} = x^{- 1} \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (x^{- 1}) \\ = - 1 x^{- 2} \\ = - x^{- 2} \\ Thus, second derivative of given function is - x^{- 2} . \end{array}$

Q.104

$\begin{array}{l} Find the second order derivative of the function : \\ x^{3} logx \end{array}$

Ans.

$\begin{array}{l} Let y = x^{3} logx \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{3} logx) \\ = x^{3} \frac{d}{dx} logx + logx \frac{d}{dx} x^{3} [By product rule] \\ = x^{3} \times \frac{1}{x} + logx \times 3 x^{2} \\ = x^{2} + 3 x^{2} logx \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (x^{2} + 3 x^{2} logx) \\ = \frac{d}{dx} x^{2} + 3 \frac{d}{dx} (x^{2} logx) \\ = 2 x + 3 (x^{2} \frac{d}{dx} logx + logx \frac{d}{dx} x^{2}) \\ = 2 x + 3 (x^{2} \times \frac{1}{x} + logx . 2 x) \\ = 2 x + 3 x + 6 xlogx \\ = 5 x + 6 xlogx \\ Thus, second derivative of given function is x (5 + 6 logx) . \end{array}$

Q.105

$\begin{array}{l} Find the second order derivative of the function : \\ e^{x} \sin 5 x \end{array}$

Ans.

$\begin{array}{l} Let y = e^{x} \sin 5 x \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (e^{x} \sin 5 x) \\ = e^{x} \frac{d}{dx} \sin 5 x + \sin 5 x \frac{d}{dx} e^{x} [By product rule] \\ = e^{x} (5 \cos 5 x) + \sin 5 x . e^{x} \\ = e^{x} (5 \cos 5 x + \sin 5 x) \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} \{e^{x} (5 \cos 5 x + \sin 5 x)\} \\ = e^{x} \frac{d}{dx} (5 \cos 5 x + \sin 5 x) + (5 \cos 5 x + \sin 5 x) \frac{d}{dx} e^{x} \\ [By product rule] \\ = e^{x} (- 25 \sin 5 x + 5 \cos 5 x) + (5 \cos 5 x + \sin 5 x) e^{x} \\ = e^{x} (- 25 \sin 5 x + 5 \cos 5 x + 5 \cos 5 x + \sin 5 x) \\ = 2 e^{x} (- 12 \sin 5 x + 5 \cos 5 x) \end{array}$

Q.106

$\begin{array}{l} Find the second order derivative of the function : \\ e^{6 x} \cos 3 x \end{array}$

Ans.

$\begin{array}{l} Let y = e^{6 x} \cos 3 x \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (e^{6 x} \cos 3 x) \\ = e^{6 x} \frac{d}{dx} \cos 3 x + \cos 3 x \frac{d}{dx} e^{6 x} [By product rule] \\ = e^{6 x} (- 3 \sin 3 x) + \cos 3 x . 6 e^{6 x} \\ = e^{6 x} (6 \cos 3 x - 3 \sin 3 x) \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} {e^{6 x} (6 \cos 3 x - 3 \sin 3 x)} \\ = e^{6 x} \frac{d}{dx} (6 \cos 3 x - 3 \sin 3 x) + (6 \cos 3 x - 3 \sin 3 x) \frac{d}{dx} e^{6 x} \\ [By product rule] \\ = e^{x} (- 18 \sin 3 x - 9 \cos 3 x) + (6 \cos 3 x - 3 \sin 3 x) 6 e^{6 x} \\ = e^{6 x} (- 18 \sin 3 x - 9 \cos 3 x + 36 \cos 3 x - 18 \sin 3 x) \\ = e^{6 x} (- 36 \sin 3 x + 27 \cos 3 x) \\ = 9 e^{6 x} (- 4 \sin 3 x + 3 \cos 3 x) \\ Thus, second derivative of given function is \\ 9 e^{6 x} (3 \cos 3 x - 4 \sin 3 x) . \end{array}$

Q.107

$\begin{array}{l} Find the second order derivative of the function : \\ \tan^{- 1} x \end{array}$

Ans.

$\begin{array}{l} Let y = \tan^{- 1} x \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \tan^{- 1} x \\ = \frac{1}{1 + x^{2}} \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (\frac{1}{1 + x^{2}}) \\ = \frac{d}{dx} {(1 + x^{2})}^{- 1} \\ = - {(1 + x^{2})}^{- 2} \frac{d}{dx} (1 + x^{2}) \\ = - {(1 + x^{2})}^{- 2} (0 + 2 x) \\ = - \frac{2 x}{{(1 + x^{2})}^{2}} \\ Thus, second derivative of given function is - \frac{2 x}{{(1 + x^{2})}^{2}} . \end{array}$

Q.108

$\begin{array}{l} Find the scond orderderivative of the function : \\ \log (logx) \end{array}$

Ans.

$\begin{array}{l} Let y = \log (logx) \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \log (logx) \\ = \frac{1}{logx} \frac{d}{dx} logx [By chain rule] \\ = \frac{1}{logx} (\frac{1}{x}) \\ = \frac{1}{xlogx} \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (\frac{1}{xlogx}) \\ = \frac{d}{dx} {(xlogx)}^{- 1} \\ = - {(xlogx)}^{- 2} \frac{d}{dx} (xlogx) [By chain rule] \\ = - \frac{1}{{(xlogx)}^{2}} (x \frac{d}{dx} logx + logx \frac{d}{dx} x) \\ = - \frac{1}{{(xlogx)}^{2}} (x \times \frac{1}{x} + logx \times 1) \\ = - \frac{1}{{(xlogx)}^{2}} (1 + logx) \\ = - \frac{(1 + logx)}{{(xlogx)}^{2}} \\ Thus, second derivative of given function is - \frac{(1 + logx)}{{(xlogx)}^{2}} . \end{array}$

Q.109 Find the second order derivative of the function:
sin (log x)

Ans.

$\begin{array}{l} Let y = \sin (logx) \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \sin (logx) \\ = \cos (logx) \frac{d}{dx} logx [By chain rule] \\ = \cos (logx) (\frac{1}{x}) \\ = \frac{\cos (logx)}{x} \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (\frac{\cos (logx)}{x}) \\ = \frac{x \frac{d}{dx} \cos (logx) - \cos (logx) \frac{d}{dx} x}{x^{2}} \\ = \frac{x \times - \sin (logx) \frac{d}{dx} logx - \cos (logx) \times 1}{x^{2}} [By chain rule] \\ = \frac{- xsin (logx) \times \frac{1}{x} - \cos (logx)}{x^{2}} \\ = \frac{- {\sin (logx) + \cos (logx)}}{x^{2}} \\ Thus, second derivative of given function is \\ \frac{- {\sin (logx) + \cos (logx)}}{x^{2}} . \end{array}$

Q.110

$If y = 5 cosx - 3 sinx, prove that \frac{d^{2} y}{d x^{2}} + y = 0$

Ans.

$\begin{array}{l} Given : \\ y = 5 cosx - 3 sinx \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (5 cosx - 3 sinx) \\ = 5 \frac{d}{dx} cosx - 3 \frac{d}{dx} sinx \\ = - 5 sinx - 3 cosx \\ Again differentiating w . r . t . x, we get \\ \frac{d^{2} y}{{dx}^{2}} = - \frac{d}{dx} (5 sinx + 3 cosx) \\ = - (5 \frac{d}{dx} sinx + 3 \frac{d}{dx} cosx) \\ = - (5 cosx - 3 sinx) \\ = - y \\ ∴ \frac{d^{2} y}{{dx}^{2}} + y = 0 \\ Hence Proved . \end{array}$

Q.111

$If y = co s^{- 1} x . Find \frac{d^{2} y}{d x^{2}} in term s o f y alone .$

Ans.

$\begin{array}{l} Given : \\ y = \cos^{- 1} x \\ cosy = x \\ Differentiating w . r . t . y, we get \\ \frac{d}{dy} cosy = \frac{d}{dy} x \\ \Rightarrow \frac{dx}{dy} = - siny \\ \Rightarrow \frac{dy}{dx} = - \frac{1}{siny} \\ \Rightarrow \frac{dy}{dx} = - cosecy . .. (i) \\ Again differentiating w . r . t . x, we get \\ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (- cosecy) \\ = cosecycoty \frac{dy}{dx} \\ = cosecycoty \times - cosecy \\ = - {cosec}^{2} ycoty [From equation (i)] \end{array}$

Q.112

$If y = 3 \cos (logx) + 4 \sin (logx), show that x^{2} y_{2} + x y_{1} + y = 0$

Ans.

$\begin{array}{l} Given : \\ y = 3 \cos (logx) + 4 \sin (logx) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} (= y_{1}) = \frac{d}{dx} {3 \cos (logx) +4 \sin (logx)} \\ = 3 \frac{d}{dx} \cos (logx) + 4 \frac{d}{dx} \sin (logx) \\ = - 3 \sin (logx) \frac{d}{dx} logx + 4 \cos (logx) \frac{d}{dx} logx [By chain rule] \\ = - 3 \sin (logx) \frac{1}{x} + 4 \cos (logx) \frac{1}{x} \\ = \frac{- 3 \sin (logx) + 4 \cos (logx)}{x} \\ Again differentiating w . r . t . x, we get \\ \frac{d^{2} y}{{dx}^{2}} (y_{2}) = \frac{d}{dx} (\frac{4 \cos (logx) - 3 \sin (logx)}{x}) \\ = \frac{x \frac{d}{dx} {4 \cos (logx) - 3 \sin (logx)} - {4 \cos (logx) - 3 \sin (logx)} \frac{d}{dx} x}{x^{2}} \\ = \frac{x {- 4 \sin (logx) \frac{d}{dx} logx - 3 \cos (logx) \frac{d}{dx} logx} - {4 \cos (logx) - 3 \sin (logx)} \times 1}{x^{2}} \\ = \frac{x {- 4 \sin (logx) \frac{1}{x} - 3 \cos (logx) \frac{1}{x}} - {4 \cos (logx) - 3 \sin (logx)}}{x^{2}} \\ = \frac{- 4 \sin (logx) - 3 \cos (logx) - 4 \cos (logx) + 3 \sin (logx)}{x^{2}} \\ = \frac{- \sin (logx) - 7 \cos (logx)}{x^{2}} \\ Thus, \\ \frac{d^{2} y}{{dx}^{2}} = y_{2} = \frac{- \sin (logx) - 7 \cos (logx)}{x^{2}} \\ L . H . S . = x^{2} y_{2} + {xy}_{1} + y \\ = x^{2} (\frac{- \sin (logx) - 7 \cos (logx)}{x^{2}}) + x (\frac{- 3 \sin (logx) + 4 \cos (logx)}{x}) \\ + 3 \cos (logx) + 4 \sin (logx) \\ = - \sin (logx) - 7 \cos (logx) - 3 \sin (logx) + 4 \cos (logx) + 3 \cos (logx) \\ + 4 \sin (logx) \\ = - 4 \sin (logx) + 4 \sin (logx) - 7 \cos (logx) + 7 \cos (logx) \\ = 0 = R . H . S . \\ Hence proved . \end{array}$

Q.113

$If y = A e^{mx} + B e^{nx}, show that \frac{d^{2} y}{d x^{2}} - (m + n) \frac{dy}{dx} + mny = 0$

Ans.

$\begin{array}{l} G i v e n t h a t \\ y = A e^{m x} + B e^{n x} \\ D i f f e r e n t i a t i n g w . r . t . x, w e g e t \\ \frac{d y}{d x} = A \frac{d}{d x} e^{m x} + B \frac{d}{d x} e^{n x} \\ = A m e^{m x} + B n e^{n x} \\ A g a i n d i f f e r e n t i a t i n g w . r . t . x, w e g e t \\ \frac{d^{2} y}{d x^{2}} = A m \frac{d}{d x} e^{m x} + B n \frac{d}{d x} e^{n x} \\ = A m^{2} e^{m x} + B n^{2} e^{n x} \\ L . H . S . = \frac{d^{2} y}{d x^{2}} - (m + n) \frac{d y}{d x} + m n y \\ = (A m^{2} e^{m x} + B n^{2} e^{n x}) - (m + n) (A m e^{m x} + B n e^{n x}) + m n (A e^{m x} + B e^{n x}) \\ = A m^{2} e^{m x} + B n^{2} e^{n x} - A m^{2} e^{m x} - B m n e^{n x} - A m n e^{m x} - B n^{2} e^{n x} \\ + A m n e^{m x} + B m n e^{n x} \\ = 0 = R . H . S . \end{array}$

Q.114

$If y = 500 e^{7 x} + 600 e^{- 7 x}, show that \frac{d^{2} y}{d x^{2}} = 49 y$

Ans.

$\begin{array}{l} Given that \\ y = 500 e^{7 x} + 600 e^{- 7 x} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = 500 \frac{d}{dx} e^{7 x} + 600 \frac{d}{dx} e^{- 7 x} \\ = 500 \times 7 e^{7 x} + 600 \times (- 7) e^{- 7 x} \\ = 7 (500 e^{7 x} - 600 e^{- 7 x}) \\ Again differentiating w . r . t . x, we get \\ \frac{d^{2} y}{{dx}^{2}} = 7 (500 \times 7 e^{7 x} - 600 \times (- 7) e^{- 7 x}) \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 49 (500 e^{7 x} + 600 e^{- 7 x}) \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 49 y \\ Hence proved . \end{array}$

Q.15

$If e^{y} (x + 1) = 1, show that \frac{d^{2} y}{d x^{2}} = {(\frac{dy}{dx})}^{2}$

Ans.

$\begin{array}{l} G i v e n that \\ e^{y} (x + 1) = 1 \\ e^{y} = \frac{1}{x + 1} \\ y = \log (\frac{1}{x + 1}) \\ y = - \log (x + 1) \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} (- \log (x + 1)) \\ = - \frac{d}{d x} \log (x + 1) \\ = - \frac{1}{x + 1} \frac{d}{d x} (x + 1) \\ = - \frac{1}{x + 1} \times (1 + 0) \\ = - \frac{1}{x + 1} \\ A g a i n d i f f e r e n t i a t i n g w .r .t . x, we get \\ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (- \frac{1}{x + 1}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{d}{d x} {(x + 1)}^{- 1} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = - (- 1) {(x + 1)}^{- 2} \frac{d}{d x} (x + 1) \\ = \frac{1}{{(x + 1)}^{2}} \\ = {(- \frac{1}{x + 1})}^{2} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x})}^{2} \\ H e n c e proved . \end{array}$

Q.116

$If y = {(ta n^{- 1} x)}^{2}, show that (x^{2} + 1) y_{2} + 2 x (x^{2} + 1) y_{1} = 2$

Ans.

$\begin{array}{l} The given function is \\ y = {(\tan^{- 1} x)}^{2} \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} {(\tan^{- 1} x)}^{2} \\ = 2 \tan^{- 1} x \frac{d}{dx} \tan^{- 1} x \\ y_{1} = 2 \tan^{- 1} x (\frac{1}{1 + x^{2}}) \\ (1 + x^{2}) y_{1} = 2 \tan^{- 1} x \\ Differentiating again w . r . t . x, we get \\ \frac{d}{dx} {(1 + x^{2}) y_{1}} = 2 \frac{d}{dx} \tan^{- 1} x \\ (1 + x^{2}) \frac{d}{dx} y_{1} + y_{1} \frac{d}{dx} (1 + x^{2}) = 2 \times \frac{1}{1 + x^{2}} \\ (1 + x^{2}) y_{2} + y_{1} (0 + 2 x) = 2 \times \frac{1}{1 + x^{2}} \\ (1 + x^{2}) y_{2} + 2 {xy}_{1} = \frac{2}{1 + x^{2}} \\ {(1 + x^{2})}^{2} y_{2} + 2 x (1 + x^{2}) y_{1} = 2 \\ \Rightarrow {(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2 \\ Hence proved . \end{array}$

Q.117

$\begin{array}{l} VerifyRollesTheoremforthe function \\ f (x) = x^{2} + 2 x - 8, x \in [- 4, 2] . \end{array}$

Ans.

$\begin{array}{l} The given function, f (x) = x^{2} + 2 x - 8, is a polynomial . So, it \\ is continuous in [- 4, 2] and is differentiable in (- 4, 2) . \\ f (- 4) = {(- 4)}^{2} + 2 (- 4) - 8 \\ = 16 - 8 - 8 = 0 \\ f (2) = {(2)}^{2} + 2 (2) - 8 \\ = 4 - 4 - 8 = 0 \\ ∴ f (- 4) = f (2) = 0 \\ \Rightarrow The value of f (x) at - 4 and 2 coincide . \\ Rolle ‘ s Theorem states that there is a point c \in (- 4, 2) such \\ that f ‘ (c) = 0 \\ f (x) = x^{2} + 2 x - 8 \\ Differentiating f (x) w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} (x^{2} + 2 x - 8) \\ = 2 x + 2 \\ f ‘ (c) = 2 c + 2 \\ âˆµ f ‘ (c) = 0 \Rightarrow 2 c + 2 = 0 \\ \Rightarrow c = - 1 \in (- 4, 2) \\ Hence, Rolle ‘ s Theorem is verified for the given function . \end{array}$

Q.118

$\begin{array}{l} Examine if Rolle ‘ s Theorem is applicable to any of the \\ following functions . Can you say some thing about the \\ converse of Rolle ‘ s theorem from these example ? \\ (i) f (x) = [x] for x \in [5,9] \\ (ii) f (x) = [x] for x \in [- 2,2] \\ (iii) f (x) = x^{2} - 1 for x \in [1,2] \end{array}$

Ans.

$\begin{array}{l} (i) Given function is f (x) = [x] for x \in [5, 9] \\ Since the given function f (x) is not continuous at every \\ integral point and in particular it is not continuous at x = 5 \\ and x = 9 . \\ \Rightarrow f (x) is not continuous in [5, 9] . \\ f (5) = [5] = 5 \\ f (9) = [9] = 9 \\ ∴ f (5) \neq f (9) \\ The differentiability of f in (5, 9) is checked as follows . \\ Let n be an integer such that n \in (5, 9) . \\ The left hand limit of f at x = n is \\ \lim_{h \to 0^{-}} \frac{f (n + h) - f (n)}{h} = \lim_{h \to 0^{-}} \frac{[n + h] - [n]}{h} \\ = \lim_{h \to 0^{-}} \frac{n - 1 - n}{h} [\begin{array}{l} Since, h is very small number . \\ So, \lim_{h \to 0^{-}} [n + h] = \lim_{h \to 0^{-}} (n - 1) \\ Example : [5 - 0 . 3] = [4 . 7] = 4 \end{array}] \\ = \lim_{h \to 0^{-}} \frac{- 1}{h} = - \infty \\ The right hand limit of f at x = n is \\ \lim_{h \to 0^{+}} \frac{f (n + h) - f (n)}{h} = \lim_{h \to 0^{+}} \frac{[n + h] - [n]}{h} \\ = \lim_{h \to 0^{+}} \frac{n - n}{h} [\begin{array}{l} Since, h is very small number . \\ So, \lim_{h \to 0^{+}} [n + h] = \lim_{h \to 0 +} (n) \\ Example : [5 + 0 .3] = [5 .3] = 5 \end{array}] \\ = \lim_{h \to 0^{+}} 0 \\ = 0 \\ ∴ Left Hand Derivative \neq Right Hand Derivative \\ So, f is not differentiable at x = n . \\ ∴ f is not differentiable in (5, 9) . \\ It is observed that f does not satisfy all the conditions of the \\ hypothesis of Roll ‘ s Theorem . \\ Hence, Roll ‘ s Theorem is not applicable for f (x) = [x] for x \in [5, 9] \\ (ii) Given function is f (x) = [x] for x \in [- 2, 2] \\ Since the given function f (x) is not continuous at every \\ integral point and in particular it is not continuous at x = - 2 \\ and x = 2 . \\ \Rightarrow f (x) is not continuous in [- 2, 2]. \\ f (- 2) = [- 2] = - 2 \\ f (2) = [2] = 2 \\ ∴ f (- 2) \neq f (2) \\ The differentiability of f in (- 2, 2) is checked as follows . \\ Let n be an integer such that n \in (- 2, 2) . \\ The left hand limit of f at x = n is \\ \lim_{h \to 0^{-}} \frac{f (n + h) - f (n)}{h} = \lim_{h \to 0^{-}} \frac{[n + h] - [n]}{h} \\ = \lim_{h \to 0^{-}} \frac{n - 1 - n}{h} [\begin{array}{l} Since, h is very small number . \\ So, \lim_{h \to 0^{-}} [n + h] = \lim_{h \to 0^{-}} (n - 1) \\ Example : [5 - 0 .3] = [4 .7] = 4 \end{array}] \\ = \lim_{h \to 0^{-}} \frac{- 1}{h} = - \infty \end{array}$ $\begin{array}{l} T h e right hand limit of f at x = n is \\ \lim_{h \to 0^{+}} \frac{f (n + h) - f (n)}{h} = \lim_{h \to 0^{+}} \frac{[n + h] - [n]}{h} \\ = \lim_{h \to 0^{+}} \frac{n - n}{h} [\begin{array}{l} S i n c e, h is very small number . \\ S o, \lim_{h \to 0^{+}} [n + h] = \lim_{h \to 0 +} (n) \\ E x a m p l e : [5 + 0.3] = [5.3] = 5 \end{array}] \\ = \lim_{h \to 0^{+}} 0 \\ = 0 \\ ∴ Left Hand Derivative \neq Right Hand Derivative \\ So, f is not differentiable at x = n . \\ ∴ f is not differentiable in (- 2, 2) . \\ It is observed that f does not satisfy all the conditions of the \\ hypothesis of Roll’s Theorem . \\ Hence, Roll’s Theorem is not applicable for f (x) = [x] for x \in [- 2, 2] \\ (i i i) T h e given function, f (x) = x^{2} - 1, is a polynomial . So, it \\ is continuous in [1, 2] and is differentiable in (1, 2) . \\ f (1) = {(1)}^{2} - 1 \\ = 0 \\ f (2) = {(2)}^{2} - 1 \\ = 4 - 1 \\ = 3 \\ ∴ f (1) \neq f (2) \\ \Rightarrow T h e values of f (x) at 1 and 2 do not coincide . \\ It is observed that f does not satisfy a condition of the \\ hypothesis of Rolle’s Theorem . \\ Hence, Rolle’s Theorem is not applicable for \\ f (x) = x^{2} - 1 f o r x \in [1, 2] \end{array}$

Q.119

$\begin{array}{l} If f : [- 5, 5] \to R is differentiable function and if f ‘ (x) does not \\ vanish anywhere, then prove that f (- 5) \neq f (5) \end{array}$

Ans.

$\begin{array}{l} Given function f : [- 5, 5] \to R ‹ is a differentiable function . \\ Since every differentiable function is a continuous function then \\ we have \\ (a) f is continuous on [- 5, 5] . \\ (b) f is differentiable on (- 5, 5) . \\ Therefore, by the Mean Value Theorem, there exists \\ c \in (- 5, 5) such that \\ f ‘ (c) = \frac{f (5) - f (- 5)}{5 - (- 5)} \\ = \frac{f (5) - f (- 5)}{10} \\ 10 f ‘ (c) = f (5) - f (- 5) \\ It is also given that f ‘ (x) does not vanish anywhere . \\ ∴ f ‘ (c) \neq 0 \\ \Rightarrow 10 f ‘ (c) \neq 0 \\ f (5) - f (- 5) \neq 0 \\ \Rightarrow f (5) \neq f (- 5) \\ Hence, proved . \end{array}$

Q.120

$\begin{array}{l} Verify Mean Value Theorem, if f (x) = x^{2} - 4 x - 3 in the \\ interval [a, b], where a = 1 and b = 4 . \end{array}$

Ans.

$\begin{array}{l} Verify Mean Value Theorem, if f (x) = x^{2} - 4 x - 3 in the \\ interval [a, b], where a = 1 and b = 4 . \end{array}$

Q.121

$\begin{array}{l} Verify Mean Value Theorem, if f (x) = x^{3} - 5 x^{2} - 3 x in the \\ interval [a, b], wherea = 1 and b = 3 . Find all c \in (1, 3) \\ for which f ‘(c) = 0 . \end{array}$

Ans.

$\begin{array}{l} The given function = x^{3} - 5 x^{2} - 3 x \\ Since, function f (x) is polynomial, so it is continuous in [1, 3] \\ and it is differentiable in (1,3) . \\ Here, f ‘ (x) = \frac{d}{dx} (x^{3} - 5 x^{2} - 3 x) \\ = 3 x^{2} - 10 x - 3 \\ and, f ‘ (c) = 3 c^{2} - 10 c - 3 \\ Now, f (1) = {(1)}^{3} - 5 {(1)}^{2} - 3 (1) \\ = 1 - 5 - 3 \\ = - 7 \\ and f (3) = {(3)}^{3} - 5 {(3)}^{2} - 3 (3) \\ = 27 - 45 - 9 \\ = - 27 \\ ∴ \frac{f (b) - f (a)}{b - a} = \frac{f (3) - f (1)}{3 - 1} \\ \Rightarrow \frac{f (b) - f (a)}{b - a} = \frac{- 27 - (- 7)}{2} \\ \Rightarrow \frac{f (b) - f (a)}{b - a} = \frac{- 20}{2} \\ \Rightarrow \frac{f (b) - f (a)}{b - a} = - 10 \\ Mean Value Theorem states that there is a point c \in (1, 3) \\ such that f ‘ (c) = - 10 \\ âˆµ f ‘ (c) = - 10 \\ \Rightarrow 3 c^{2} - 10 c + 7 = 0 \\ \Rightarrow 3 c^{2} - 3 c - 7 c + 7 = 0 \\ \Rightarrow 3 c (c - 1) c - 7 (c - 1) = 0 \\ \Rightarrow (c - 1) (3 c - 7) = 0 \\ \Rightarrow c = 1, \frac{7}{3}, where c = \frac{7}{3} \in (1, 3) \\ Hence, Mean Value Theorem is verified for the given function \\ c = \frac{7}{3} \in (1, 3) is the only point for which f ‘ (c) = 0 . \end{array}$

Q.122

$\begin{array}{l} Examine the applicability of Mean Value Theorem for all three \\ functions given below . \\ (i) f (x) = [x] f o r x \in [5, 9] \\ (i i) f (x) = [x] f o r x \in [- 2, 2] \\ (i i i) f (x) = x^{2} - 1 f o r x \in [1, 2] \end{array}$

Ans.

$\begin{array}{l} (i) Given function is f (x) = [x] for x \in [5, 9] \\ Since the given function f (x) is not continuous at every \\ integral point and in particular it is not continuous at x = 5 \\ and x = 9 . \\ \Rightarrow f (x) is not continuous in [5, 9] . \\ f (5) = [5] = 5 \\ f (9) = [9] = 9 \\ ∴ f (5) \neq f (9) \\ The differentiability of f in (5, 9) is checked as follows . \\ Let n be an integer such that n \in (5, 9) . \\ The left hand limit of f at x = n is \\ \lim_{h \to 0^{-}} \frac{f (n + h) - f (n)}{h} = \lim_{h \to 0^{-}} \frac{[n + h] - [n]}{h} \\ = \lim_{h \to 0^{-}} \frac{n - 1 - n}{h} [\begin{array}{l} Since, h is very small number . \\ So, \lim_{h \to 0^{-}} [n + h] = \lim_{h \to 0^{-}} (n - 1) \\ Example : [5 - 0 .3] = [4 .7] = 4 \end{array}] \\ = \lim_{h \to 0^{-}} \frac{- 1}{h} = - \infty \\ The right hand limit of f at x = n is \\ \lim_{h \to 0^{+}} \frac{f (n + h) - f (n)}{h} = \lim_{h \to 0^{+}} \frac{[n + h] - [n]}{h} \\ = \lim_{h \to 0^{+}} \frac{n - n}{h} [\begin{array}{l} Since, h is very small number . \\ So, \lim_{h \to 0^{+}} [n + h] = \lim_{h \to 0 +} (n) \\ Example : [5 + 0 .3] = [5 .3] = 5 \end{array}] \\ = \lim_{h \to 0^{+}} 0 \\ = 0 \\ ∴ Left Hand Derivative \neq Right Hand Derivative \\ So, f is not differentiable at x = n . \\ ∴ f is not differentiable in (5, 9) . \\ It is observed that f does not satisfy all the conditions of the \\ hypothesis of Mean Value Theorem . \\ Hence, Mean Value Theorem is not applicable for \\ f (x) = [x] for x \in [5, 9] \\ (ii) Given function is f (x) = [x] for x \in [- 2, 2] \\ Since the given function f (x) is not continuous at every \\ integral point and in particular it is not continuous at x = - 2 \\ and x = 2 . \\ \Rightarrow f (x) is not continuous in [- 2, 2]. \\ The differentiability of f in (- 2, 2) is checked as follows . \\ Let n be an integer such that n \in (- 2, 2) . \\ The left hand limit of f at x = n is \\ \lim_{h \to 0^{-}} \frac{f (n + h) - f (n)}{h} = \lim_{h \to 0^{-}} \frac{[n + h] - [n]}{h} \\ = \lim_{h \to 0^{-}} \frac{n - 1 - n}{h} [\begin{array}{l} Since, h is very small number . \\ So, \lim_{h \to 0^{-}} [n + h] = \lim_{h \to 0^{-}} (n - 1) \\ Example : [5 - 0 .3] = [4 .7] = 4 \end{array}] \\ = \lim_{h \to 0^{-}} \frac{- 1}{h} = - \infty \\ The right hand limit of f at x = n is \\ \lim_{h \to 0^{+}} \frac{f (n + h) - f (n)}{h} = \lim_{h \to 0^{+}} \frac{[n + h] - [n]}{h} \\ = \lim_{h \to 0^{+}} \frac{n - n}{h} [\begin{array}{l} Since, h is very small number . \\ So, \lim_{h \to 0^{+}} [n + h] = \lim_{h \to 0 +} (n) \\ Example : [5 + 0 .3] = [5 .3] = 5 \end{array}] \\ = \lim_{h \to 0^{+}} 0 \\ = 0 \\ ∴ Left Hand Derivative \neq Right Hand Derivative \\ So, f is not differentiable at x = n . \\ ∴ f is not differentiable in (- 2, 2) . \\ It is observed that f does not satisfy all the conditions of the \\ hypothesis of Mean Value Theorem . \\ Hence, Mean Value Theorem is not applicable for \\ f (x) = [x] for x \in [- 2, 2] \end{array}$ $\begin{array}{l} (i i i) T h e given function, f (x) = x^{2} - 1, is a polynomial . So, it \\ is continuous in [1, 2] and is differentiable in (1, 2) . \\ It is observed that f satisfies all the conditions of the hypothesis \\ of Mean Value Theorem . \\ Hence, Mean Value Theorem is applicable for \\ f (x) = x^{2} - 1 f o r x \in [1, 2] \\ It can be proved as: \\ f (1) = {(1)}^{2} - 1 = 0 \\ f (2) = {(2)}^{2} - 1 = 3 \\ ∴ \frac{f (b) - f (a)}{b - a} = \frac{f (2) - f (1)}{2 - 1} \\ = \frac{3 - 0}{1} \\ = 3 \\ âˆµ f ‘ (x) = 2 x \\ ∴ f ‘ (c) = 3 \\ \Rightarrow 2 c = 3 \\ \Rightarrow c = \frac{3}{2} = 1.5 \in [1, 2] \end{array}$

Q.123 Differentiate w.r.t. x the function:
(3x² – 9x + 5)⁹.

Ans.

$\begin{array}{l} Let y = {(3 x^{2} - 9 x + 5)}^{9} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} {(3 x^{2} - 9 x + 5)}^{9} \\ = 9 {(3 x^{2} - 9 x + 5)}^{8} \frac{d}{dx} (3 x^{2} - 9 x + 5) [By chain rule] \\ = 9 {(3 x^{2} - 9 x + 5)}^{8} (6 x - 9) \\ = 9 \times 3 (2 x - 3) {(3 x^{2} - 9 x + 5)}^{8} \\ = 27 (2 x - 3) {(3 x^{2} - 9 x + 5)}^{8} \end{array}$

Q.124 Differentiate w.r.t. x the function:
sin³x + cos⁶x

Ans.

Differentiate w.r.t. x the function:
sin³x + cos⁶x

Q.125 Differentiate w.r.t. x the function: (5 x )^3cos2x

Ans.

$\begin{array}{l} Let y = {(5 x)}^{3 \cos 2 x} \\ Taking \log both sides, we get \\ logy = \log {(5 x)}^{3 \cos 2 x} \\ logy = 3 \cos 2 x . \log 5 x \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} logy = 3 \cos 2 x \frac{d}{dx} \log 5 x + 3 \log 5 x \frac{d}{dx} \cos 2 x \\ \frac{1}{y} \frac{dy}{dx} = 3 \cos 2 x (\frac{1}{5 x} \frac{d}{dx} 5 x) + 3 \log 5 x (- \sin 2 x) \frac{d}{dx} 2 x [By chain rule] \\ \frac{dy}{dx} = y {\frac{3}{5 x} \cos 2 x (5) - 3 \log 5 xsin 2 x (2)} \\ = {(5 x)}^{3 \cos 2 x} {\frac{3 \cos 2 x}{x} - 6 \sin 2 x \log 5 x} \end{array}$

Q.126

$\begin{array}{l} D i f f e r e n t i a t e w . r . t . x t h e f u n c t i o n : \\ s i n^{- 1} (x \sqrt{x}), 0 \leq x \leq 1 \end{array}$

Ans.

$\begin{array}{l} Let y = \sin^{- 1} (x \sqrt{x}) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \sin^{- 1} (x \sqrt{x}) \\ = \frac{1}{\sqrt{1 - {(x \sqrt{x})}^{2}}} \frac{d}{dx} (x \sqrt{x}) [By chain rule] \\ = \frac{1}{\sqrt{1 - x^{3}}} \frac{d}{dx} x^{\frac{3}{2}} \\ = \frac{1}{\sqrt{1 - x^{3}}} \times \frac{3}{2} x^{\frac{1}{2}} \\ = \frac{3 \sqrt{x}}{2 \sqrt{1 - x^{3}}} \\ = \frac{3}{2} \sqrt{\frac{x}{1 - x^{3}}} \end{array}$

Q.127

$\begin{array}{l} D i f f e r e n t i a t e w . r . t . x t h e f u n c t i o n : \\ \frac{c o s^{- 1} (\frac{x}{2})}{\sqrt{2 x + 7}}, - 2 < x < 2 \end{array}$

Ans.

$\begin{array}{l} Lety = \frac{\cos^{- 1} (\frac{x}{2})}{\sqrt{2 x + 7}}, - 2 < x < 2 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (\frac{\cos^{- 1} (\frac{x}{2})}{\sqrt{2 x + 7}}) \\ = \frac{\sqrt{2 x + 7} \frac{d}{dx} \cos^{- 1} (\frac{x}{2}) - \cos^{- 1} (\frac{x}{2}) \frac{d}{dx} \sqrt{2 x + 7}}{{(\sqrt{2 x + 7})}^{2}} [By Quotientrule] \\ = \frac{\sqrt{2 x + 7} (- \frac{1}{\sqrt{1 - {(\frac{x}{2})}^{2}}}) \frac{d}{dx} (\frac{x}{2}) - \cos^{- 1} (\frac{x}{2}) (\frac{1}{2 \sqrt{2 x + 7}}) \frac{d}{dx} (2 x + 7)}{2 x + 7} [By chainrule] \\ = \frac{- \frac{2 \sqrt{2 x + 7}}{\sqrt{4 - x^{2}}} \times \frac{1}{2} - \cos^{- 1} (\frac{x}{2}) (\frac{1}{2 \sqrt{2 x + 7}}) (2)}{2 x + 7} \\ \frac{dy}{dx} = \frac{- \frac{\sqrt{2 x + 7}}{\sqrt{4 - x^{2}}} - \cos^{- 1} (\frac{x}{2}) (\frac{1}{\sqrt{2 x + 7}})}{2 x + 7} \\ = - \frac{1}{\sqrt{4 - x^{2}}} \times \frac{1}{\sqrt{2 x + 7}} - \frac{\cos^{- 1} (\frac{x}{2})}{{(2 x + 7)}^{\frac{3}{2}}} \\ \frac{dy}{dx} = - (\frac{1}{\sqrt{4 - x^{2}}} \times \frac{1}{\sqrt{2 x + 7}} + \frac{\cos^{- 1} (\frac{x}{2})}{{(2 x + 7)}^{\frac{3}{2}}}) \end{array}$

Q.128

$\begin{array}{l} D i f f e r e n t i a t e w . r . t . x t h e f u n c t i o n : \\ c o t^{- 1} [\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}}], 0 \leq x \leq \frac{π}{2} \end{array}$

Ans.

$\begin{array}{l} Let \cot^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}], 0 \leq x \leq \frac{π}{2} \\ Now, \frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}} \\ = \frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}} \times \frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} + \sqrt{1 - sinx}} \\ = \frac{{(\sqrt{1 + sinx} + \sqrt{1 - sinx})}^{2}}{{(\sqrt{1 + sinx})}^{2} - {(\sqrt{1 - sinx})}^{2}} \\ = \frac{1 + sinx + 1 - sinx + 2 \sqrt{1 + sinx} \sqrt{1 - sinx}}{1 + sinx - (1 - sinx)} \\ = \frac{2 + 2 \sqrt{1 - \sin^{2} x}}{1 + sinx - 1 + sinx} \\ = \frac{2 + 2 cosx}{2 sinx} \\ = \frac{2 (1 + cosx)}{2 sinx} \\ = \frac{2 \cos^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\ = \cot \frac{x}{2} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \cot^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}] \\ = \frac{d}{dx} \cot^{- 1} [\cot \frac{x}{2}] [By Quotient rule] \\ = \frac{d}{dx} (\frac{x}{2}) \\ = \frac{1}{2} . \\ Thus, ‹ \frac{dy}{dx} = \frac{1}{2} . \end{array}$

‘Q.129

$\begin{array}{l} Differentiate w . r . t . x the function : \\ {(logx)}^{logx}, x > 1 \end{array}$

Ans.

$\begin{array}{l} Let y = {(logx)}^{logx}, x > 1 \\ Taking \log both sides, we get \\ logy = \log {(logx)}^{logx} \\ logy = logx . \log (logx) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} logy = logx \frac{d}{dx} \log (logx) + \log (logx) \frac{d}{dx} logx [By product rule] \\ \frac{1}{y} \frac{dy}{dx} = logx (\frac{1}{logx} \frac{d}{dx} logx) + \log (logx) \frac{1}{x} [By chain rule] \\ \frac{dy}{dx} = y \{\frac{1}{x} + \frac{1}{x} \log (logx)\} \\ = {(logx)}^{logx} \{\frac{1}{x} + \frac{1}{x} \log (logx)\} \end{array}$

Q.130

$\begin{array}{l} Differentiate w . r . t . x the function : \\ \cos (a \cos x + b \sin x), for some constant a and b . \end{array}$

Ans.

$\begin{array}{l} Let y = \cos (acosx + bsinx) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \cos (acosx + bsinx) \\ = - \sin (acosx + bsinx) \frac{d}{dx} (acosx + bsinx) [By chain rule] \\ = - \sin (acosx + bsinx) \times (- asinx + bcosx) \\ = - (bcosx - asinx) \sin (acosx + bsinx) \end{array}$

Q.131

$\begin{array}{l} Differentiate w . r . t . x the function : \\ {(sinx - cosx)}^{(sinx - cosx)}, \frac{π}{4} < x < \frac{3 π}{4} \end{array}$

Ans.

$\begin{array}{l} Let y = {(sinx - cosx)}^{(sinx - cosx)}, \frac{π}{4} < x < \frac{3 π}{4} \\ Taking \log both sides, we get \\ logy = \log {(sinx - cosx)}^{(sinx - cosx)} \\ logy = (sinx - cosx) . \log (sinx - cosx) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} logy = (sinx - cosx) \frac{d}{dx} \log (sinx - cosx) \\ + \log (sinx - cosx) \frac{d}{dx} (sinx - cosx) [By product rule] \\ \frac{1}{y} \frac{dy}{dx} = (sinx - cosx) (\frac{1}{(sinx - cosx)} \frac{d}{dx} (sinx - cosx)) \\ + \log (sinx - cosx) . (cosx + sinx) [By chain rule] \\ \frac{dy}{dx} = y [\begin{array}{l} (sinx - cosx) {\frac{1}{(sinx - cosx)} (cosx + sinx)} \\ + (cosx + sinx) \log (sinx - cosx) \end{array}] \\ \frac{dy}{dx} = {(sinx - cosx)}^{(sinx - cosx)} (cosx + sinx) {1 + \log (sinx - cosx)} \end{array}$

Q.132

$\begin{array}{l} Differentiate w . r . t . x the function : \\ x^{x} + x^{a} + a^{x} + a^{a}, for some fixed a > 0 and x > 0 \end{array}$

Ans.

$\begin{array}{l} Let y = x^{x} + x^{a} + a^{x} + a^{a} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{x} + x^{a} + a^{x} + a^{a}) \\ = \frac{d}{dx} x^{x} + \frac{d}{dx} x^{a} + \frac{d}{dx} a^{x} + \frac{d}{dx} a^{a} \\ = \frac{d}{dx} x^{x} + {ax}^{a - 1} + a^{x} \log_{e} a + 0 \\ = \frac{d}{dx} x^{x} + {ax}^{a - 1} + a^{x} \log_{e} a . .. (i) \\ Let u = x^{x} \\ Taking \log both sides, we get \\ logu = {logx}^{x} \\ = xlogx \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} logu = \frac{d}{dx} (xlogx) \\ \frac{1}{u} \frac{du}{dx} = x \frac{d}{dx} logx + logx \frac{d}{dx} x \\ \frac{du}{dx} = u {x . \frac{1}{x} + logx . (1)} \\ \frac{du}{dx} = x^{x} (1 + logx) \\ From equation (i), we get \\ \frac{dy}{dx} = x^{x} (1 + logx) + {ax}^{a - 1} + a^{x} \log_{e} a \end{array}$

Q.133

$\begin{array}{l} Differentiate w . r . t . x the function : \\ x^{x^{2} - 3} + {(x - 3)}^{x^{2}}, for x > 3 \end{array}$

Ans.

$\begin{array}{l} Let y = x^{x^{2} - 3} + {(x - 3)}^{x^{2}} \\ Let u = x^{x^{2} - 3}, v = {(x - 3)}^{x^{2}} \\ So, y = u + v \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} . .. (i) \\ Taking \log both sides, we get \\ logu = {logx}^{x^{2} - 3} \\ = (x^{2} - 3) logx \\ and \\ logv = \log {(x - 3)}^{x^{2}} \\ = x^{2} \log (x - 3) \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} logu = \frac{d}{dx} {(x^{2} - 3) logx} \\ \frac{1}{u} \frac{du}{dx} = (x^{2} - 3) \frac{d}{dx} logx + logx \frac{d}{dx} (x^{2} - 3) \\ \frac{du}{dx} = u {(x^{2} - 3) . \frac{1}{x} + logx . (2 x)} \\ \frac{du}{dx} = x^{(x^{2} - 3)} {\frac{(x^{2} - 3)}{x} + logx . (2 x)} \\ and \\ \frac{d}{dx} logv = \frac{d}{dx} {x^{2} \log (x - 3)} \\ \frac{1}{v} \frac{dv}{dx} = x^{2} \frac{d}{dx} \log (x - 3) + \log (x - 3) \frac{d}{dx} x^{2} \\ \frac{dv}{dx} = v {x^{2} . \frac{1}{x - 3} + \log (x - 3) . (2 x)} \\ \frac{dv}{dx} = {(x - 3)}^{x^{2}} {\frac{x^{2}}{x - 3} + 2 x . \log (x - 3)} \\ From equation (i), we get \\ \frac{dy}{dx} = x^{(x^{2} - 3)} {\frac{(x^{2} - 3)}{x} + 2 x . logx} + {(x - 3)}^{x^{2}} {\frac{x^{2}}{x - 3} + 2 x . \log (x - 3)} \end{array}$

Q.134

$Find \frac{dy}{dx}, if y = 12 (1 - \cos t), x = 10 (t - \sin t), - \frac{π}{2} < t < \frac{π}{2}$

Ans.

$\begin{array}{l} It is given that, y = 12 (1 - cost), x = 10 (t - sint) \\ Differentiating w . r . t . t, we get \\ \frac{dy}{dt} = \frac{d}{dt} 12 (1 - cost) and \frac{dx}{dt} = \frac{d}{dt} 10 (t - sint) \\ \frac{dy}{dt} = 12 (0 + sint) and \frac{dx}{dt} = 10 (1 - cost) \\ \frac{dy}{dt} = 12 sint and \frac{dx}{dt} = 10 (1 - cost) \\ ∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ = \frac{12 sint}{10 (1 - cost)} \\ \Rightarrow \frac{dy}{dx} = \frac{6 \times 2 \sin \frac{t}{2} \cos \frac{t}{2}}{5 \times 2 \sin^{2} \frac{t}{2}} \\ \Rightarrow \frac{dy}{dx} = \frac{6}{5} \cot \frac{t}{2} . \end{array}$

Q.135

$Find \frac{dy}{dx}, if y = si n^{- 1} x + si n^{- 1} \sqrt{1 - x^{2}}, - 1 \leq x \leq 1$

Ans.

$\begin{array}{l} Let y = \sin^{- 1} x + \sin^{- 1} \sqrt{1 - x^{2}}, - 1 \leq x \leq 1 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (\sin^{- 1} x + \sin^{- 1} \sqrt{1 - x^{2}}) \\ = \frac{d}{dx} \sin^{- 1} x + \frac{d}{dx} \sin^{- 1} \sqrt{1 - x^{2}} \\ = \frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{\sqrt{1 - {(\sqrt{1 - x^{2}})}^{2}}} \frac{d}{dx} (\sqrt{1 - x^{2}}) [By chain rule] \\ = \frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{\sqrt{1 - (1 - x^{2})}} \times \frac{1}{2 \sqrt{1 - x^{2}}} \frac{d}{dx} (1 - x^{2}) \\ = \frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{\sqrt{x^{2}}} \times \frac{1}{2 \sqrt{1 - x^{2}}} (0 - 2 x) \\ = \frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{x} \times \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x) \\ = \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - x^{2}}} \\ ∴ \frac{dy}{dx} = 0 . \end{array}$

Q.136

$\begin{array}{l} If x \sqrt{1 + y} + y \sqrt{1 + x} = 0, for, - 1 < x < 1, prove that \\ \frac{dy}{dx} = - \frac{1}{{(1 + x)}^{2}} \end{array}$

Ans.

$\begin{array}{l} Given equation is x \sqrt{1 + y} + y \sqrt{1 + x} = 0 \\ x \sqrt{1 + y} = - y \sqrt{1 + x} \\ Squarring both sides, we get \\ x^{2} (1 + y) = y^{2} (1 + x) \\ \Rightarrow x^{2} + x^{2} y = y^{2} + {xy}^{2} \\ \Rightarrow x^{2} - y^{2} = {xy}^{2} - x^{2} y \\ \Rightarrow (x - y) (x + y) = xy (y - x) \\ \Rightarrow (x - y) (x + y) = - xy (x - y) \\ \Rightarrow (x + y) = - xy \\ \Rightarrow y + xy = - x \\ \Rightarrow (1 + x) y = - x \\ \Rightarrow y = - \frac{x}{1 + x} \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = - \frac{d}{dx} \frac{x}{(1 + x)} \\ \Rightarrow \frac{dy}{dx} = - {\frac{(1 + x) \frac{d}{dx} x - x \frac{d}{dx} (1 + x)}{{(1 + x)}^{2}}} \\ \Rightarrow \frac{dy}{dx} = - {\frac{(1 + x) . 1 - x (0 + 1)}{{(1 + x)}^{2}}} \\ \Rightarrow \frac{dy}{dx} = - {\frac{1 + x - x}{{(1 + x)}^{2}}} \\ \Rightarrow \frac{dy}{dx} = - \frac{1}{{(1 + x)}^{2}} \\ Hence proved . \end{array}$

Q.137

$\begin{array}{l} If {(x - a)}^{2} + {(y - b)}^{2} = c^{2}, for some c > 0, prove that \\ \frac{{[1 + {(\frac{dy}{dx})}^{2}]}^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} is a constant independent of a and b . \end{array}$

Ans.

$\begin{array}{l} Given equation is {(x - a)}^{2} + {(y - b)}^{2} = c^{2} \\ Differentiating both sides w . r . t . x, we get \\ \frac{d}{dx} {(x - a)}^{2} + \frac{d}{dx} {(y - b)}^{2} = \frac{d}{dx} c^{2} \\ \Rightarrow 2 (x - a) \frac{d}{dx} (x - a) + 2 (y - b) \frac{d}{dx} (y - b) = 0 [By chain rule] \\ \Rightarrow 2 (x - a) (1 - 0) + 2 (y - b) (\frac{dy}{dx} - 0) = 0 \\ \Rightarrow 2 (x - a) + 2 (y - b) \frac{dy}{dx} = 0 \\ \Rightarrow \frac{dy}{dx} = - \frac{x - a}{y - b} . .. (i) \\ Differentiating both sides again w . r . t . x, we get \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = - \frac{d}{dx} (\frac{x - a}{y - b}) [By quotient rule] \\ = - \frac{(y - b) \frac{d}{dx} (x - a) - (x - a) \frac{d}{dx} (y - b)}{{(y - b)}^{2}} \\ = - \frac{(y - b) (1 - 0) - (x - a) (\frac{dy}{dx} - 0)}{{(y - b)}^{2}} \\ = - \frac{(y - b) - (x - a) (- \frac{x - a}{y - b})}{{(y - b)}^{2}} [From equation (i)] \\ \frac{d^{2} y}{{dx}^{2}} = - \frac{(y - b) - (x - a) (- \frac{x - a}{y - b})}{{(y - b)}^{2}} \\ = - \frac{{(y - b)}^{2} + {(x - a)}^{2}}{{(y - b)}^{3}} \\ ∴ \frac{d^{2} y}{{dx}^{2}} = - \frac{c^{2}}{{(y - b)}^{3}} [âˆµ {(x - a)}^{2} + {(y - b)}^{2} = c^{2}] \\ Now, \frac{{[1 + {(\frac{dy}{dx})}^{2}]}^{\frac{3}{2}}}{\frac{d^{2} y}{{dx}^{2}}} = \frac{{[1 + {(- \frac{x - a}{y - b})}^{2}]}^{\frac{3}{2}}}{- \frac{c^{2}}{{(y - b)}^{3}}} \\ = \frac{{\frac{{(y - b)}^{2} + {(x - a)}^{2}}{{(y - b)}^{2}}}^{\frac{3}{2}}}{- \frac{c^{2}}{{(y - b)}^{3}}} \\ = \frac{{\frac{c^{2}}{{(y - b)}^{2}}}^{\frac{3}{2}}}{- \frac{c^{2}}{{(y - b)}^{3}}} \\ = \frac{\frac{c^{3}}{{(y - b)}^{3}}}{- \frac{c^{2}}{{(y - b)}^{3}}} \\ = - c, \\ which is a contant and is independent of a and b . \\ Hence proved . \end{array}$

Q.138

$\begin{array}{l} If \cos âŸ y âŸ = âŸ x âŸ \cos (a + y), with âŸ \cos âŸ a âŸ \neq \pm âŸ 1, âŸ prove âŸ that \\ âŸâŸ âŸâŸ âŸ \frac{dy}{dx} = \frac{\cos^{2} (a + y)}{sina} \end{array}$

Ans.

$\begin{array}{l} Given equation is \\ cosy = xcos (a + y) . .. (i) \\ Differentiating ‹ both sides with respect to x, we get \\ \frac{d}{dx} cosy = \frac{d}{dx} {xcos (a + y)} \\ - siny \frac{dy}{dx} = x \frac{d}{dx} \cos (a + y) + \cos (a + y) \frac{d}{dx} x \\ = x \times - \sin (a + y) \frac{d}{dx} (a + y) + \cos (a + y) \times 1 \\ = - xsin (a + y) \frac{dy}{dx} + \cos (a + y) \\ {xsin (a + y) - siny} \frac{dy}{dx} = \cos (a + y) \\ \Rightarrow {\frac{cosy}{\cos (a + y)} \sin (a + y) - siny} \frac{dy}{dx} = \cos (a + y) \\ [Putting value of x from equation (i)] \\ \Rightarrow {cosysin (a + y) - sinycos (a + y)} \frac{dy}{dx} = \cos^{2} (a + y) \\ \Rightarrow \sin (a + y - y) \frac{dy}{dx} = \cos^{2} (a + y) \\ \Rightarrow sina \frac{dy}{dx} = \cos^{2} (a + y) \\ \Rightarrow \frac{dy}{dx} = \frac{\cos^{2} (a + y)}{sina} \\ Hence proved . \end{array}$

Q.139

$If x = a (cost + t sint) and y = a (sint - tcost), find \frac{d^{2} y}{d x^{2}} .$

Ans.

$\begin{array}{l} Given equations are x = a (cost + t sint) and y = a (sint - t cost) \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} a (cost + t sint) \\ = a (- sint + t \frac{d}{dt} sint + sint \frac{d}{dt} t) \\ = a (- sint + tcost + sint) \\ = atcost \\ \frac{dy}{dt} = \frac{d}{dt} a (sint - t cost) \\ = a (cost - t \frac{d}{dt} cost - cost \frac{d}{dt} t) \\ = a (cost + tsint - cost) \\ = atsint \\ ∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ = \frac{atsint}{atcost} \\ = tant \\ Again differentiating w . r . t . x, we get \\ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} tant \\ = \sec^{2} t \frac{dt}{dx} \\ = \sec^{2} t \times \frac{1}{atcost} [âˆµ \frac{dx}{dt} = atcost \Rightarrow \frac{dt}{dx} = \frac{1}{atcost}] \\ = \frac{\sec^{3} t}{at}, 0 < t < \frac{π}{2} \end{array}$

Q.140

$If f (x) = {| x |}^{3}, show that f ” (x) exists for all realx, and find it .$

Ans.

$\begin{array}{l} Since, | x | = {\begin{cases} x, if x \geq 0 \\ - x, if x < 0 \end{cases} \\ Therefore, when x \geq 0, f (x) = {| x |}^{3} = x^{3} \\ So, f ‘ (x) = 3 x^{2} and f ” (x) = 6 x \\ when x < 0, f (x) = {| x |}^{3} = {(- x)}^{3} = - x^{3} \\ So, f ‘ (x) = - 3 x^{2} and f ” (x) = - 6 x \\ Thus, for f (x) = {| x |}^{3}, \\ f ‘ ‘ (x) exists for all real x and is given by, \\ f ‘ ‘ (x) = {\begin{cases} 6 x, if x \geq 0 \\ - 6 x, if x <0 \end{cases} \end{array}$

Q.141

$\begin{array}{l} Using mathematical induction prove that \frac{d}{dx} (x^{n}) = n x^{n - 1} for \\ all positive integers n . \end{array}$

Ans.

$\begin{array}{l} Let P (n) : \frac{d}{dx} (x^{n}) = {nx}^{n - 1} for all positive integers n \\ For n = 1, \\ P (1) : \frac{d}{dx} (x^{1}) = 1 = 1 x^{1 - 1} \\ ∴ P (n) is true for n = 1 \\ Let P (k) is true for some positive integer k . \\ P (k) : \frac{d}{dx} (x^{k}) = {kx}^{k - 1} . .. (i) \\ It should be true for n = k +1 \\ So, \\ P (k + 1) : \frac{d}{dx} (x^{k + 1}) = (k + 1) x^{(k + 1) - 1} \\ \Rightarrow P (k + 1) : \frac{d}{dx} (x^{k + 1}) = (k + 1) x^{k} \\ L . H . S . = \frac{d}{dx} (x^{k + 1}) \\ = \frac{d}{dx} (x^{k} . x) \\ = x^{k} \frac{d}{dx} x + x \frac{d}{dx} x^{k} \\ = x^{k} + x . {kx}^{k - 1} \\ = x^{k} + {kx}^{k} \\ = (1 + k) x^{k} \\ = (1 + k) x^{(k + 1) - 1} = R . H . S . \\ Thus, P (k + 1) is true whenever P (k) is true . \\ Therefore, by the principle of mathematical induction, the \\ statement P (n) is true for every positive integer n . \\ Hence proved . \end{array}$

Q.142 Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines. <strong>Ans.

$\begin{array}{l} Given equation is \\ \sin (A + B) = sinAcosB + cosAsinB \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} \sin (A + B) = \frac{d}{dx} (sinAcosB) + \frac{d}{dx} (cosAsinB) \\ \Rightarrow \cos (A + B) \frac{d}{dx} (A + B) = sinA \frac{d}{dx} cosB + (\frac{d}{dx} sinA) cosB \\ + cosA \frac{d}{dx} sinB + (\frac{d}{dx} cosA) sinB \\ \Rightarrow \cos (A + B) (\frac{dA}{dx} + \frac{dB}{dx}) = sinA (- sinB \frac{dB}{dx}) + (cosA \frac{dA}{dx}) cosB \\ + cosAcosB \frac{dB}{dx} + (- sinA \frac{dA}{dx}) sinB \\ \Rightarrow \cos (A + B) (\frac{dA}{dx} + \frac{dB}{dx}) = - sinAsinB \frac{dB}{dx} + cosA \frac{dA}{dx} cosB \\ + cosAcosB \frac{dB}{dx} - sinA \frac{dA}{dx} sinB \\ \Rightarrow \cos (A + B) (\frac{dA}{dx} + \frac{dB}{dx}) = (cosAcosB - sinAsinB) \frac{dA}{dx} \\ + (cosAcosB - sinAsinB) \frac{dB}{dx} \\ \Rightarrow \cos (A + B) (\frac{dA}{dx} + \frac{dB}{dx}) = (cosAcosB - sinAsinB) (\frac{dA}{dx} + \frac{dB}{dx}) \\ \Rightarrow \cos (A + B) = (cosAcosB - sinAsinB) \end{array}$

Q.143 Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Ans.

$\begin{array}{l} Yes, there exists a function which is continuous everywhere \\ but not differentiable at exactly two points . \\ For example : \\ f (x) = | x - 1 | + | x - 2 |, x \in R \\ is continuous at everywhere but not differentiable at x =1 and 2 . \\ f (x) = | x - 1 | + | x - 2 | \\ = {\begin{cases} - (x - 1) - (x - 2), if x < 1 \\ (x - 1) - (x - 2), if 1 \leq x < 2 \\ (x - 1) + (x - 2), if x \geq 2 \end{cases} \\ = {\begin{cases} - 2 x + 3, if x < 1 \\ 1, if 1 \leq x < 2 \\ 2 x - 3, if x \geq 2 \end{cases} \\ Since, f is continuous at all points . \\ So, ‹ differentiability of f at x = 1 \\ L . H . D . = \lim_{h \to 0^{-}} \frac{f (1 + h) - f (1)}{h} \\ = \lim_{h \to 0^{-}} \frac{- 2 (1 + h) + 3 - {- 2 (1) + 3}}{h} \\ = \lim_{h \to 0^{-}} \frac{- 2 - 2 h + 3 + 2 - 3}{h} \\ = \lim_{h \to 0^{-}} \frac{- 2 h}{h} \\ = - 2 \\ R . H . D . = \lim_{h \to 0^{+}} \frac{f (1 + h) - f (1)}{h} \\ = \lim_{h \to 0^{+}} \frac{1 - 1}{h} \\ = \lim_{h \to 0^{+}} \frac{0}{h} \\ = 0 \\ ∴ L . H . D . \neq R . H . D . \\ Thus, function is not differentiable at x = 1. \\ Now, ‹ differentiability of f at x = 2 \\ L . H . D . = \lim_{h \to 0^{-}} \frac{f (2 + h) - f (2)}{h} \\ = \lim_{h \to 0^{-}} \frac{1 - 1}{h} \\ = \lim_{h \to 0^{-}} \frac{0}{h} \\ = 0 \\ R . H . D . = \lim_{h \to 0^{+}} \frac{f (2 + h) - f (2)}{h} \\ = \lim_{h \to 0 +} \frac{2 (2 + h) - 3 - 1}{h} \\ = \lim_{h \to 0^{+}} \frac{4 + 2 h - 4}{h} \\ = 2 \\ ∴ L . H . D . \neq R . H . D . \\ Thus, function f is not differentiable at x = 2. \\ Therefore, function f is continuous at everywhere but it is not \\ differentiable at exactly two points x =1 and 2 . \end{array}$

Q.144

$If y = | \begin{array}{l} f (x) g (x) h (x) \\ l m n \\ a b c \end{array} |, p r o v e t h a t \frac{d y}{d x} = | \begin{array}{l} f^{'} (x) g^{'} (x) h^{'} (x) \\ l m n \\ a b c \end{array} |$

Ans.

$\begin{array}{l} W e h a v e \\ y = | \begin{array}{l} f (x) g (x) h (x) \\ l m n \\ a b c \end{array} | \\ = f (x) | \begin{array}{l} m n \\ b c \end{array} | + g (x) | \begin{array}{l} n l \\ c a \end{array} | + h (x) | \begin{array}{l} l m \\ a b \end{array} | \\ D i f f e r e n t i a t i n g w i t h r e s p e c t t o x, w e g e t \\ \frac{d y}{d x} = \frac{d}{d x} f (x) | \begin{array}{l} m n \\ b c \end{array} | + \frac{d}{d x} g (x) | \begin{array}{l} n l \\ c a \end{array} | + \frac{d}{d x} h (x) | \begin{array}{l} l m \\ a b \end{array} | \\ \frac{d y}{d x} = f^{'} (x) | \begin{array}{l} m n \\ b c \end{array} | + g^{'} (x) | \begin{array}{l} n l \\ c a \end{array} | + h^{'} (x) | \begin{array}{l} l m \\ a b \end{array} | \\ = | \begin{array}{l} f^{'} (x) g^{'} (x) h^{'} (x) \\ l m n \\ a b c \end{array} | \\ T h e r e f o r e, \\ \frac{d y}{d x} = | \begin{array}{l} f^{'} (x) g^{'} (x) h^{'} (x) \\ l m n \\ a b c \end{array} | \end{array}$

Q.145

$\begin{array}{l} If y = e^{{acos}^{- 1} x}, - 1 \leq x \leq 1, show that \\ (1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - x \frac{dy}{dx} - a^{2} y = 0 . \end{array}$

Ans.

$\begin{array}{l} Given that, y = e^{{acos}^{- 1} x} \\ Taking logarithm both sides, we get \\ logy = {loge}^{{acos}^{- 1} x} \\ = {acos}^{- 1} x loge \\ = {acos}^{- 1} x \times 1 [âˆµ loge = 1] \\ Differentiating both sides, with respect to x, we get \\ \frac{d}{dx} logy = \frac{d}{dx} {acos}^{- 1} x \\ \Rightarrow \frac{1}{y} \frac{dy}{dx} = a \frac{- 1}{\sqrt{1 - x^{2}}} \\ \Rightarrow \frac{dy}{dx} = - \frac{ay}{\sqrt{1 - x^{2}}} \\ Squarring both sides, we get \\ {(\frac{dy}{dx})}^{2} = {(- \frac{ay}{\sqrt{1 - x^{2}}})}^{2} \\ \Rightarrow {(\frac{dy}{dx})}^{2} = \frac{a^{2} y^{2}}{(1 - x^{2})} \\ \Rightarrow (1 - x^{2}) {(\frac{dy}{dx})}^{2} = a^{2} y^{2} \\ Differentiating again both sides, with respect to x, we get \\ (1 - x^{2}) \frac{d}{dx} {(\frac{dy}{dx})}^{2} + {(\frac{dy}{dx})}^{2} \frac{d}{dx} (1 - x^{2}) = a^{2} \frac{d}{dx} y^{2} \\ (1 - x^{2}) . 2 (\frac{dy}{dx}) \frac{d^{2} y}{{dx}^{2}} - 2 x {(\frac{dy}{dx})}^{2} = 2 a^{2} y \frac{dy}{dx} \\ 2 (\frac{dy}{dx}) {(1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - x (\frac{dy}{dx})} = 2 a^{2} y \frac{dy}{dx} \\ \Rightarrow (1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - x (\frac{dy}{dx}) = a^{2} y [âˆµ \frac{dy}{dx} \neq 0] \\ \Rightarrow (1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - x (\frac{dy}{dx}) - a^{2} y = 0 \\ Hence proved . \end{array}$

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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NCERT Solutions Class 12 Mathematics Chapter 5 covers all important topics. It includes continuity, differentiability, exponential and logarithmic functions, logarithmic differentiation, and second-order derivatives. Further, the students will also understand the core concepts of the mean value theorem.

The NCERT Solutions Class 12 Mathematics Chapter 5 have been designed by Extramarks Subject Matter Experts. Every question is covered with simplified and step-by-step solutions. It plays a crucial role in preparing for competitive exams such as JEE Main and JEE Advanced.

There are eight exercises in NCERT Solutions for Class 12 Mathematics Chapter 5. These are

Exercise 5.1 – 34 questions
Exercise 5.2 – 10 questions
Exercise 5.3 – 15 questions
Exercise 5.4 – 10 questions
Exercise 5.5 – 18 questions
Exercise 5.6 – 11 questions
Exercise 5.7 – 17 questions
Exercise 5.8 – 6 questions
Miscellaneous exercise – 23 questions

NCERT Solutions Class 12 Mathematics Chapter 5 – Continuity and Differentiability

Key Topics Covered In NCERT Solutions for Class 12 Mathematics Chapter 5

5.1 Introduction

5.2 Continuity

5.3 Differentiability

5.4 Exponential and Logarithmic Functions

5.5 Logarithmic Differentiation

5.6 Derivatives of Functions in Parametric Forms

5.7 Second Order Derivative

5.8 Mean Value Theorem

NCERT Solutions Class 12 Mathematics Chapter 5 Exercise & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features of NCERT Solutions Class 12 Mathematics Chapter 5

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

FAQs (Frequently Asked Questions)

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NCERT Solutions Class 12 Mathematics Chapter 5 – Continuity and Differentiability

Key Topics Covered In NCERT Solutions for Class 12 Mathematics Chapter 5

5.1 Introduction

5.2 Continuity

5.3 Differentiability

5.4 Exponential and Logarithmic Functions

5.5 Logarithmic Differentiation

5.6 Derivatives of Functions in Parametric Forms

5.7 Second Order Derivative

5.8 Mean Value Theorem

NCERT Solutions Class 12 Mathematics Chapter 5 Exercise & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features of NCERT Solutions Class 12 Mathematics Chapter 5

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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