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

Prove that the function  fx=5x3  is continuous at x=0,at x=3 and at x=5.

Ans.

Givenfunctionis  f(x)=5x3At x=0,f(0)=5(0)3=3limx0f(x)=limx05x3      =5(0)3      =3limx0f(x)=f(0)Therefore, f is continuous at x = 0.At x=3,f(3)=5(3)3=18limx3f(x)=limx05x3      =5(3)3      =18limx3f(x)=f(3)Therefore, f is continuous at x = 3.At x=5, f(5)=5(5)3=22limx5f(x)=limx05x3      =5(5)3      =22limx5f(x)=f(5)Therefore, f is continuous at x = 5.

 

Q.2

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

Ans.

The given function is f(x)=2x21At x=3,f(3)=2(3)21=17limx3f(x)=limx32x21      =2(3)21      =17limx3f(x)=f(3)Therefore, f is continuous at x = 3.

 

Q.3

Examine the following functions for continuity.a fx=x5bfx=1x5,  x3c fx=x225x+5,x5dfx=x5

Ans.

(a)  The given function is  f(x)=x5It is evident that f is defined at every real number k, then f(k) = k 5.  limxkf(x)=limxk(x5)=k5limxkf(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)=1x5,x5For any real number k5 then f(k) = k 5.  limxkf(x)=limxk(x5)=k5limxkf(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)=x225x+5,x5For any real number k5 then     f(k)=(k5)(k+5)(k+5)=k 5  limxkf(x)=limxk(k5)(k+5)(k+5))=k5limxkf(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|={x+5,ifx<5x5,ifx5This 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 > 5Case I: c < 5Then, f (c) = 5climxcf(x)=limxc(5x)      =5climxcf(x)=f (c)Therefore, f is continuous at all real numbers less than 5.Case II: c=5Then, f (c)=f (5)    =55=0      limx5f(x)=limx5(5x)    =55=0      limx5+f(x)=limx5(x5)    =55=0limx5f(x)=limx5+f(x)=f (c)Therefore, f is continuous at x = 5Case III: c > 5Then,  f (c) = c5limxcf(x)=limxc(x5)=c5limxcf(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.

 

Q.4

Prove that the function f x = x n  is continuous at x = n, where n is a positive integer.

Ans.

The given function is f (x) = xnIt is evident that f is defined at all positive integers, n, and its value at n is nn.Then,limxnf(n)=limxnxn=nnlimxnf(x)=f(n)Therefore, f is continuous at n, where n is a positive integer.

 

Q.5

Is the function f defined byfx=x,  if  x15,  if  x>1continuous at x=0? Atx=1? Atx=2?

Ans.

The given function f isf(x)={x,  if  x15,  if  x>1At x=0It is evident that f is defined at 0 and its value at 0 is 0.Then, limx0f(x)=limx0x=0limx0f(x)=f(0)Therefore, f is continuous at x = 0At x = 1,f is defined at 1 and its value at 1 is 1.The left hand limit of f at x = 1 is,limx1f(x)=limx1x=1The right hand limit of f at x = 1 is,limx1+f(x)=limx1+(5)=5limx1f(x)limx1+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,limx2f(x)=limx2(5)=5limx2f(x)=f(2)Therefore, f is continuous at x = 2.

 

Q.6

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

f(x)={2x+3,  if  x22x3,  if  x>2

Ans.

The given function f isf(x)={2x+3,ifx22x3,ifx>2It 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(ii)c>2(iii)c=2Case1:c<2Then,f(c)=2c+3limxcf(x)=limxc(2x+3)=2c+3limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x < 2.Case2:c>2Then,f(c)=2c3limxcf(x)=limxc(2x3)=2c3limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x > 2.Case (iii) c = 2Then, the left hand limit of f at x = 2 is,limx2f(x)=limx2(2x+3)=7limx2+f(x)=limx2(2x3)=1limx2f(x)limx2+f(x)Therefore, f is not continuous at x = 2Hence, x = 2 is the only point of discontinuity of f.

 

Q.7

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

f(x)={|x|+3,  if  x32x,  if  3<x<36x+2,ifx3

Ans.

The given function f is  f(x)={|x|+3=x+3,    if  x32x,          if3 < x < 36x+3,        if  x3The given function f is defined at all the points of the real line.Let c be a point on the real line. Case1:Ifc<3Then,  f(c)=c+3limxcf(x)=limxc(x+3)=c+3limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x <3.Case2:c=3Then, f(3)=(3)+3=6limx3f(x)=limx3(x+3)=(3)+3=6limx3+f(x)=limx3+(2x)=2(3)=6limx3f(x)=limx3+f(x)=f(3)Therefore, f is continuous at x = 3.Case3:If3<c<3,then      f(c)=2c  andlimxcf(x)=limxc(2x)=2climxcf(x)=f(c)Therefore, f is continuous in (3,3).Case 4:If c=3, thenL.H.L.=limx3f(x)=limx3(2x)=2(3)=6R.H.L.=limx3+f(x)=limx3+(6x+2)=6(3)+2=20limx3f(x)=limx3+f(x)Therefore, f is not continuous at x = 3Case 5:If c>3, thenf(c)=6c+2 and limxcf(x)=limxc(6x+2)=6c+2limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x > 3Hence, x = 3 is the only point of discontinuity of f.

 

Q.8

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

f(x)={|x|x,  if  x00,  if  x=0

Ans.

The given function f is f(x)={|x|x, if  x00,  if  x=0={|x|x=xx=1,  if  x>00,  if  x=0|x|x=xx=1,  if  x<0The 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<0, then f(c)=1  limxcf(x)=limxc(1)=1limxcf(x)=f(c)Therefore, f is continuous at all points x < 0.CaseII:If c=0, then L.H.L. at x=0 is,L.H.L.=limx0f(x)=limx0(1)=1L.H.L. at x=0 ​isR.H.L.=limx0+f(x)=limx0+(1)=1L.H.L.R.H.L.Therefore, f is not continuous at x = 0CaseIII:If c>0,thenf(c)=1limxcf(x)=limxc(1)=1limxcf(x)=f(c)Therefore, f is continuous at all points x > 0.Hence, x = 0 is the only point of discontinuity of f.

 

Q.9

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

f(x)={x|x|,  if  x<01,  if  x0

Ans.

The given function f is f(x)={x|x|=xx=1,  if  x<01,  if  x0f(x)=1 for all xRLet c be any real number. Then,limxcf(x)=limxc(1)=1Andf(c)=1for all xRTherefore, the given function is a continuous function.Hence, the given function has no point of discontinuity.

 

Q.10

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

f(x)={x+1,  if  x1x2+1,  if  x<1

Ans.

The given function f is f(x)={x+1,  if  x1x2+1,  if  x<1The 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<1, then f(c)=c2+1​​​ andlimxcf(x)=limxc(x2+1)      =c2+1limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x < 1.CaseII:If c=1, then f(c)=f(1)=1+1=2The left hand limit of f at x = 1 is,limx1f(x)=limx1(x2+1)=12+1=2The right hand limit of f at x = 1 is,limx1+f(x)=limx1+(x+1)=1+1=2limx1f(x)=f(1)Therefore, f is continuous at x = 1.CaseIII:If c>1, then f(c)=c+1limxcf(x)=limxc(x+1)      =c+1limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x > 1Hence, the given function f has no point of discontinuity.

 

Q.11

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

f(x)={x33,  if  x2x2+1,  if  x>2

Ans.

The given function f is f(x)={x33,  if  x2x2+1,  if  x>2The 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<1, then f(c)=c2+1​​​ andlimxcf(x)=limxc(x2+1)      =c2+1limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x < 1.CaseII:If c=1, then f(c)=f(1)=1+1=2The left hand limit of f at x = 1 is,limx1f(x)=limx1(x2+1)=12+1=2The right hand limit of f at x = 1 is,limx1+f(x)=limx1+(x+1)=1+1=2limx1f(x)=f(1)Therefore, f is continuous at x = 1.CaseIII:If c>1, then f(c)=c+1limxcf(x)=limxc(x+1)      =c+1limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x > 1Hence, the given function f has no point of discontinuity.The given function f is f(x)={x+1,  if  x1x2+1,  if  x<1The 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<1, then f(c)=c2+1​​​ andlimxcf(x)=limxc(x2+1)      =c2+1limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x < 1.CaseII:If c=1, then f(c)=f(1)=1+1=2The left hand limit of f at x = 1 is,limx1f(x)=limx1(x2+1)=12+1=2The right hand limit of f at x = 1 is,limx1+f(x)=limx1+(x+1)=1+1=2limx1f(x)=f(1)Therefore, f is continuous at x = 1.CaseIII:If c>1, then f(c)=c+1limxcf(x)=limxc(x+1)      =c+1limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x > 1Hence, the given function f has no point of discontinuity.

 

Q.12

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

f(x)={x101,  if  x1x2,  if  x>1

Ans.

The given function f is f(x)={x101,if  x1x2, if  x<1The 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<1, then f(c)=c101​​​ andlimxcf(x)=limxc(x101)      =c101limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x < 1.CaseII:If c=1, The left hand limit of f at x = 1 is,limx1f(x)=limx1(x101)=1101=0The right hand limit of f at x = 1 is,limx1+f(x)=limx1+(x2)=12=1limx1f(x)limx1+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 = 1CaseIII:If c>1, then f(c)=c2limxcf(x)=limxc(x2)      =c2limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x > 1Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

 

Q.13

Is the function defined by f(x)=x+5if x1x5if x >1a continuous function?

Ans.

The given function f is f(x)={x+5,if  x1x5,if  x>1The 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<1, then f(c)=c+5​​​ andlimxcf(x)=limxc(x+5)      =c+5limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x < 1.CaseII:If c=1, then f(1)=1+5=6The left hand limit of f at x = 1 is,limx1f(x)=limx1(x+5)=1+5=6The right hand limit of f at x = 1 is,limx1+f(x)=limx1+(x5)=15=4limx1f(x)limx1+f(x)Therefore, f is not continuous at x = 1CaseIII:If c>1, then f(c)=c5  andlimxcf(x)=limxc(x5)      =c5limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x > 1Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

 

Q.14

Discuss the continuity of the function f, where f is defined by

f(x)={3,  if  0x14,  if  1<x<35,      if  3x10

Ans.

The given function f is f ( x ) = { 3 , i f 0 x 1 4 , i f 1 < x < 3 5 , i f 3 x 10 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 c < 1 , t h e n f ( c ) = 3 a n d lim x c ⁡ f ( x ) = lim x c ( 3 ) = 3 lim x c ⁡ 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 1 ⁡ f ( x ) = lim x 1 ( 3 ) = 3 The right hand limit of f at x = 1 is, lim x 1 + ⁡ f ( x ) = lim x 1 + ( 4 ) = 4 lim x 1 ⁡ f ( x ) lim x 1 + ⁡ 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 c ⁡ f ( x ) = lim x c ( 4 ) = 4 lim x c ⁡ 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 3 ⁡ f ( x ) = lim x 3 ( 4 ) = 4 The right hand limit of f at x = 3 is, lim x 3 + ⁡ f ( x ) = lim x 3 + ( 5 ) = 5 lim x 3 ⁡ f ( x ) lim x 3 + ⁡ f ( x ) Therefore, f is not continuous at x = 3 . C a s e I V : I f 3 < c 10 , t h e n f ( c ) = 5 a n d lim x c ⁡ f ( c ) = lim x c ( 5 ) = 5 lim x c ⁡ 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 .

 

Q.15

Discuss the continuity of the function f, where f is defined by

f(x)={2x,  if  x<00,  if0x<14x,    if  x>1

Ans.

The given function f is f(x)={2x, if  x<00, if  0x<14x,ifx>1The 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>0, then f(c)=2c​​limxcf(x)=limxc(2x)      =2climxcf(x)=f(c)Therefore, f is continuous at all points x, such that x < 0.CaseII:If c=0, then f(c)=f(0)=0The left hand limit of f at x = 0 is,limx0f(x)=limx0(2x)=2×0=0The right hand limit of f at x = 0 is,limx0+f(x)=limx0+(0)=0limx0f(x)=limx0+f(x)=f(c)Therefore, f is continuous at x = 0.CaseIII:If 0< c<1, then f(c)=0.limxcf(x)=limxc(0)=0      limxcf(x)=f(c)Therefore, f is continuous at all points of the interval (0, 1).CaseIV:If c=1, then f(c)=f(1)=0.The left hand limit of f at x = 1 is,limx1f(x)=limx1(0)=0The right hand limit of f at x = 1 is,limx1+f(x)=limx1+(4x)=4×1=4limx1f(x)limx1+f(x)Therefore, f is not continuous at x = 1.CaseV:If c<1, then f(c)=4c and    limxcf(x)=limxc(4x)=4climxcf(x)=f(c)Therefore, f is continuous at all points x, such that x < 1Hence, f is not continuous only at x = 1.

 

Q.16

Discuss the continuity of the function f, where f is defined by

f(x)={2,  if  x<x2x,  if0x<12,      if  x>1

Ans.

The given function f is f(x)={2, if  x12x, if1x<12,ifx>1The 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<1, then f(c)=2and​​limxcf(x)=limxc(2)      =2limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x <1.CaseII:If c=1, then f(c)=f(1)=2The left hand limit of f at x =1 is,limx1f(x)=limx1(2)=2The right hand limit of f at x =1 is,limx1+f(x)=limx1+(2x)=2×(1)=2limx1f(x)=f(1)Therefore, f is continuous at x =1.CaseIII:If 1< c<1, then f(c)=2c.limxcf(x)=limxc(2x)=2c        limxcf(x)=f(c)Therefore, f is continuous at all points of the interval (-1, 1).CaseIV:If c=1, then f(c)=f(1)=2×1=2.The left hand limit of f at x = 1 is,limx1f(x)=limx1(2x)=2×1=2The right hand limit of f at x = 1 is,limx1+f(x)=limx1+(2)=2limx1f(x)=f(c)Therefore, f is not continuous at x = 1.CaseV:If c>1, then f(c)=2 and    limxcf(x)=limxc(2)=2limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x > 1Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.

 

Q.17

Find the relationship between a and b so that the functionf defined byfx=ax+1,  if  x3bx+,  if  x>3 tinuous at x=3.

Ans.

The given function f is f(x)={ax+1,if  x3bx+3,if  x>3If f is continuous at x = 3, thenlimx3f(x)=limx3+f(x)=f(3)...(i)limx3f(x)=limx3(ax+1)=3a+1limx3+f(x)=limx3+(bx+3)=3b+3        f(3)=3a+1Therefore, from (1), we obtain      3a+1=3b+3=3a+13a+1=3b+3  3a=3b+31    a=b+23Therefore, the required relationship is given by  a=b+23.

 

Q.18

For what value of λ is the function f defined byfx=λx22x,  if  x04x+,  if  x>0tinuous at x=0? What about continuity at x=1?

Ans.

The given function f isf(x)={λ(x22x), if  x04x+1,  if  x>0If f is continuous at x=0, thenlimx0f(x)=limx0+f(x)=f(0)limx0λ(x22x)=limx0+4x+1=λ(022×0)    λ(022×0)=4(0)+1=00=1=0, which is not possible.Therefore, there is no value of λ for which f is continuous at x = 0At x = 1,f (1) = 4x + 1 = 4 × 1 + 1 = 5limx1(4x+1)=4×1+1=5    limx1f(x)=f(1)Therefore, for any values of λ, f is continuous at x = 1.

 

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.

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]=nn=0The left hand limit of f at x = n is,  limxng(x)=limxn(x[x])=limxnxlimxn[x]=n(n1)=1The right hand limit of f at x = n is,  limxn+g(x)=limxn+(x[x])=limxn+xlimxn+[x]=nn=0  limxng(x)limxn+g(x)Therefore, f is not continuous at x = n.Hence, g is discontinuous at all integral points.

 

Q.20

Is the function defined by f(x)=x2 sinx+5continuous at x=π?

Ans.

The given function is  f(x)=x26sinx+5∵f is defined at x = πAt​ x=π,f(π)=π26sinπ+5=π26(0)+5=π2+5limxπf(x)=limxπ(x26sinx+5)Putxπ+hIf  xπ, then h0limxπf(x)=limxπ(x26sinx+5)=limh0{(h+π)26sin(h+π)+5}=limh0(h+π)26limh0sin(h+π)+limh05=limh0(h+π)26limh0(sinhcosπ+sinπcosh)+5=(0+π)26limh0(sinh×1+0×cosh)+5=π2+6sin0+5=π2+5limxπf(x)=f(π)Therefore, the given function f is continuous at x = π.

 

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.

If two functions(f and g) are continuous then their sum, difference and product (i.e., f+g, fg 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 + hIf xc, then h0g(c)=sinc  andlimxcg(x)=limxcsinx=limh0sin(c+h)      =limh0(sinccosh+sinhcosc)      =sinccos0+sin0cosc      =sinc×1+0×cosc      =sinclimxcg(x)=g(c)Therefore, g is a continuous function.Let h (x) = cos xIt is also evident that h (x) = cos x is defined for every real number.Let c be a real number. Put x = c + hIf xc, then h0h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)      =limh0(cosccoshsinhsinc)      =cosccos0+sin0sinc      =cosc×1+0×sinc      =cosclimxch(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 xcos x is a continuous function(c) f (x) = g (x) × h (x) = sin x × cos x is a continuous function

 

Q.22

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

Ans.

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 + hIf xc, then h0g(c)=sinc  andlimxcg(x)=limxcsinx=limh0sin(c+h)      =limh0(sinccosh+sinhcosc)      =sinccos0+sin0cosc      =sinc×1+0×cosc      =sinclimxcg(x)=g(c)Therefore, g is a continuous function.Let h (x) = cos xIt is also evident that h (x)=cos x is defined for every real number.Let c be a real number. Put x=c + hIf xc, then h0      h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)      =limh0(cosccoshsinhsinc)      =cosccos0+sin0sinc      =cosc×1+0×sinc      =cosclimxch(x)=h(c)Therefore, h is a continuous function.It can be concluded that,cosecx=1sinx,sinx0  is​ continuous.cosecx,nx0(nZ)is​ continuous.Therefore, cosecant is continuous except at x = , nZ.secx=1cosx,cosx0  is​ continuous.secx,  x(2n+1)π2  (nZ)  is​ continuous.Therefore, secant is continuous except atx=(2n+1)π2  (nZ).cotx=cosxsinx,  sinx0  is​ continuous.cotx,  x  (nZ)  is​ continuous.Therefore, cotangent is continuous except at x = , nZ.

 

Q.23

Find the points of discontinuity of f, wherefx=sinxx,  if  x<0x+,  if  x0

Ans.

The given function f is f(x)={sinxx, if  x<0x+1,if  x1The 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<0, then f(c)=sincc​​​ andlimxcf(x)=limxcsinxx      =sincclimxcf(x)=f(c)Therefore, f is continuous at all points x, such that x < 0.CaseII:If c>0, then f(c)=c+1limxcf(x)=limxc(x+1)      =c+1limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x > 0.CaseIII:If c=0, then f(c)=f(0)        =0+1=1The left hand limit of f at x = 0 is,limx0f(x)=limx0sinxx=1The right hand limit of f at x = 0 is,limx0+f(x)=limx0+(x+1)=1limx0f(x)=limx0+f(x)=f(0)Therefore, f is not continuous at x=0From 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.

 

Q.24

Determine if f defined byfx=x2sinx,  if  x0,  if  x=0is a continuous function?

Ans.

The given function f is f(x)={x2sin1x, if  x00, if  x=1The given function f is defined at all the points of the real line.Let c be a point on the real line.CaseI:If c0, then f(c)=c2sin1c​​​ andlimxcf(x)=limxc(x2sin1x)      =c2sin1c​​​limxcf(x)=f(c)Therefore, f is continuous at all points x, such that x0.CaseII:If c=0, then f(0)=0limx0f(x)=limx0(x2sin1x)Since,  1sin1x1,  x0x2x2sin1xx2limx0x2limx0x2sin1xlimx0x20limx0x2sin1x0limx0x2sin1x=0limx0f(x)=0Similarly,limx0+f(x)=limx0+(x2sin1x)        =limx0x2sin1x=0limx0f(x)=f(0)=limx0+f(x)Therefore, f is continuous at x = 0From the above observations, it can be concluded that f is continuous at every point of the real line.Thus, f is a continuous function.

 

Q.25

Examine the continuity of f, where f is defined byfx=sinxcosx,  if  x0,  if  x=0

Ans.

The given function f is f(x)={sinxcosx, if  x01, if  x=1The given function f is defined at all the points of the real line.Let c be a point on the real line.CaseI: If c0, then f(c)=sinccosclimxcf(x)=limxc(sinxcosx)=sinccosclimxcf(x)=f(c)Therefore, f is continuous at all points x, such that x0.CaseII:If c=0, then f(0)=1limx0f(x)=limx0(sinxcosx)        =sin0cos0        =01=1limx0+f(x)=limx0+(sinxcosx)        =sin0cos0        =01=1limx0f(x)=limx0+f(x)=f(0)Therefore, f is continuous at x = 0From the above observations, it can be concluded that f is continuous at every point of the real line.Thus, f is a continuous function.

 

Q.26

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

f(x)={kcosxπ2x,  if  xπ23,  if  x=0    atx=π2

 

Ans.

The given function f is f(x)={kcosxπ2x, if  xπ23, if  x=π2The given function f is continuous at x=π2,if f is defined at x=π2  and if the value of the f atx=π2  equals the limit of f atx=π2.Since, f is defined at x=π2and  f(π2)=3limxπ2f(x)=limxπ2kcosxπ2xLetx=π2+h, then xπ2h0limxπ2f(x)=limxπ2kcosxπ2x=limh0kcos(π2+h)π2(π2+h)=limh0ksinh2h=klimh0sinh2h=k×12=k2limxπ2f(x)=f(π2)      k2=3      k=6Therefore, the required value of k is 6.

 

Q.27

Find the values of k so that the function f is continuous at the indicated point.fx=x2,  if  x,  if  x>2  at x =2

Ans.

The given function f is f(x)={kx2, if  x23, if  x>2The 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 andf(2)=k(2)2=4klimx2f(x)=limx2+f(x)=f(2)      limx2(kx2)=limx2+(3)=4k            k×22=3=4k    4k=3      k=34Therefore, the required value of  k  is34.

 

Q.28

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

f(x)={kx+1,  if  xπcosx,  if  x>π    atx=π

Ans.

The given function f is f(x)={kx+1, if  xπcosx, if  x>πThe 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    limxπf(x)=limxπ+f(x)=f(π)    limxπ(kx+1)=limxπ+(cosx)=4π+1                +1=cosπ=4π+1+1=1=4π+1          k=2πTherefore, the required value of  k  is2π.

 

Q.29

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

f(x)={kx+1,ifx53x5,ifx>5atx=5

Ans.

The given function f is f(x)={kx+1, if  x53x5, if  x>5The 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)=5k+1        limx5f(x)=limx5+f(x)=f(5)      limx5(kx+1)=limx5+(3x5)=5k+1                5k+1=155=5k+15k+1=10          k=95Therefore, the required value of  k  is  95.


Q.30

Find the values of a and b such that the function defined byfx=,  if  xax+b,  if<x<,if  xis a continuous function.

Ans.

The given function f is f( x )={ 5, ifx2 ax+b, if2<x<10 21, ifx10 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. So,f is continuous at x = 2 and x = 10. Since f is continuous at x = 2, we obtain lim x 2 f( x )= lim x 2 + f( x )=f( 2 ) lim x 2 ( 5 )= lim x 2 + ( ax+b )=5 5=2a+b=5 2a+b=5 ( i ) Since f is continuous at x = 10, we obtain lim x 10 f( x )= lim x 10 + f( x )=f( 10 ) lim x 10 ( ax+b )= lim x 2 + ( 21 )=21 10a+b=21=21 10a+b=21 ( ii ) On subtracting equation (i) from equation (ii), we obtain 8a=16 a= 16 8 =2 By putting a=2 in equation ( ii ), we get 10( 2 )+b=21 b=2120 =1 Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5DAC@

Q.31

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

Ans.

The given function is f (x) = cos (x2)This function f is defined for every real number and f can be written as the compositionof two functions as,f = gοh, where g(x) = cos x and h(x) = x2[∵(gοh)(x)=g(h(x))=g(x2)=cosx2]First we have to prove that g (x) = cos x and h (x) = x2 are continuous functions.It is evident that g is defined for every real number.Let c be a real number.Then, g(c) = cos cPut x=c+hIfxc,then  h0  limxcg(x)=limxccosx=limh0cos(c+h)=limh0(cosccoshsincsinh)=limh0cosccoshlimh0sincsinh=cosccos0sincsin0=cosc×1sinc×0=cosclimxcg(x)=coscTherefore, g (x) = cos x is continuous function. h(x)=x2Clearly, h is defined for every real number.Let k be a real number, then h(k)=k2  limxch(x)=limxcx2=k2  limxch(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(x2) is a continuous function.

 

Q.32

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

Ans.

The given function is f (x) = |cosx|This function f is defined for every real number and f can be written as the compositionof two functions as,f = gοh, where g(x) =|x| and h(x) = cosx[∵(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|={x,ifx<0x,ifx0Clearly, g is defined for all real numbers.Let c be a real number.CaseI:  If c<0, then g(c)=c and limxcg(x)=limxc(x)=climxcg(x)=g(c)Therefore, g is continuous at all points x, such that x < 0.CaseII:If c>0, then g(c)=cand  limxcg(x)=limxc(x)=climxcg(x)=g(c)Therefore, g is continuous at all points x, such that x > 0.CaseIII:If c=0, then g(c)=g(0)=0limx0g(x)=limx0(x)=0limx0+g(x)=limx0+(x)=0limx0g(x)=limx0+g(x)=g(0)Therefore, g is continuous at x = 0From 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 + hIfxc, then h0 h (c)=cos climxch(x)=limxccosx=limh0cos(c+h)=limh0(cosccoshsincsinh)=limh0cosccoshlimh0sincsinh=cosccos0sincsin0=cosc×1sinc×0=cosclimxch(x)=coscTherefore, 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.

 

Q.33

Examine that sin|x| is a continuous function.

Ans.

The given function is f (x) = sin|x|This function f is defined for every real number and f can be written as the compositionof two functions as,f = gοh, where g(x) =|x| and h(x) = sinx[∵(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|={x,ifx<0x,ifx0Clearly, g is defined for all real numbers.Let c be a real number.CaseI:  If c<0, then g(c)=c and limxcg(x)=limxc(x)=climxcg(x)=g(c)Therefore, g is continuous at all points x, such that x < 0.CaseII:If c>0, then g(c)=cand  limxcg(x)=limxc(x)=climxcg(x)=g(c)Therefore, g is continuous at all points x, such that x > 0.CaseIII:If c=0, then g(c)=g(0)=0limx0g(x)=limx0(x)=0limx0+g(x)=limx0+(x)=0limx0g(x)=limx0+g(x)=g(0)Therefore, g is continuous at x = 0From 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 + hIfxc, then h0 h (c)=sinclimxch(x)=limxcsinx=limh0sin(c+h)=limh0(sinccoshcoscsinh)=limh0sinccoshlimh0coscsinh=sinccos0coscsin0=sinc×1cosc×0=sinclimxch(x)=sincTherefore, 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.

 

Q.34

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

Ans.

The given function is  f(x)=|x||x+1|The two functions, g and h, are defined asg(x)=|x|  and  h(x)=|x+1|Then, f = ghThe continuity of g and h is checked first.g(x)=|x|={x,ifx<0x,ifx0Clearly, g is defined for all real numbers.Let c be a real number.CaseI:  If c<0, then g(c)=c and limxcg(x)=limxc(x)=climxcg(x)=g(c)Therefore, g is continuous at all points x, such that x < 0.CaseII:If c>0, then g(c)=cand  limxcg(x)=limxc(x)=climxcg(x)=g(c)Therefore, g is continuous at all points x, such that x > 0.CaseIII:If c=0, then g(c)=g(0)=0limx0g(x)=limx0(x)=0limx0+g(x)=limx0+(x)=0limx0g(x)=limx0+g(x)=g(0)Therefore, g is continuous at x = 0From the above three observations, it can be concluded that g is continuous at all points.h(x)=|x+1| can be written ash(x)={(x+1), if x<1x+1,ifx1Since,h is defined for every real number.Let c be a real number.CaseI:Ifc<1, then h(c)=(c+1)and limxch(x)=limxc[(x+1)]      =(c+1)limxch(x)=h(c)Therefore, h is continuous at all points x, such that x <1.CaseII:Ifc>1,thenh(c)=c+1 andlimxch(x)=limxc[(x+1)]      =(c+1)limxch(x)=h(c)Therefore, h is continuous at all points x, such that x >1CaseIII:Ifc=1,thenh(c)=h(1)=1+1=0limx1h(x)=limx1[(x+1)]=(1+1)=0limx1+h(x)=limx1+[(x+1)]=(1+1)=0limx1h(x)=limx1+h(x)=h(1)Therefore, h is continuous at x = 1From the above three observations, it canbe concluded that his continuous at all points of the real line.g and h are continuous functions. Therefore, f = gh is also a continuous function.Therefore, f has no point of discontinuity.

Q.35

Differentiate the function with respect to x.

sin(x2 + 5)

Ans.

Let​ f(x)=sin(x2+5),  u(x)=x2+5 and v(t)=sint.Then(vοu)(x)=v(u(x))=v(x2+5)=sin(x2+5)=f(x)Thus, f is composite of two functions. Put t= u(x)=x2+5, then dvdt=ddtsint=cost=cos(x2+5)dtdx=ddx(x2+5)=2x+0=2xHence,by chain ruledfdx=dvdt.dtdx      =cos(x2+5).2x      =2xcos(x2+5)ddxsin(x2+5)=2xcos(x2+5)

Q.36

Differentiate the function with respect to x.

cos(sinx)

Ans.

Let  y=cos(sinx)Differentiating w.r.t. x, we getdydx=ddx[cos(sinx)]      =sin(sinx).ddxsinx[Bychain  rule]      =sin(sinx).cosx      =cosx.sin(sinx)ddxcos(sinx)=cosx.sin(sinx)

Q.37

Differentiate the function with respect to x.

sin (ax+b)

Ans.

Lety=sin(ax+b)Differentiating w.r.t. x, we getdydx=ddx[sin(ax+b)]      =cos(ax+b).ddx(ax+b)[Bychain  rule]      =cos(ax+b).a      =acos(ax+b)ddxsin(ax+b)=acos(ax+b)

Q.38

Differentiate the function with respect to x.

sec(tan(x))

Ans.

Let y=sec(tan(x))Differentiating w.r.t. x, we getdydx=ddxsec(tan(x))      =sec(tan(x))tan(tan(x))ddxtan(x)[∵ddxsecx=secxtanx]      =sec(tan(x))tan(tan(x)).sec2(x)ddx(x)[∵ddxtanx=sec2x]      =sec(tan(x))tan(tan(x)).sec2(x).12xddxsec(tan(x))=12xsec(tanx)tan(tanx).sec2(x)

Q.39

Differentiate the functions with respect to x sin(ax+b)cos(cx+d)

Ans.

Let y=sin(ax+b)cos(cx+d)Differentiating w.r.t. x, we getdydx=ddx{sin(ax+b)cos(cx+d)}      =cos(cx+d)ddxsin(ax+b)sin(ax+b)ddxcos(cx+d)cos2(cx+d)[By Quotient rule]      =cos(cx+d)cos(ax+b)ddx(ax+b)sin(ax+b)×sin(cx+d)ddx(cx+d)cos2(cx+d)      =acos(cx+d)cos(ax+b)+csin(ax+b)sin(cx+d)cos2(cx+d)      =acos(ax+b)cos(cx+d)+csin(ax+b)sin(cx+d)cos2(cx+d)      =acos(ax+b)sec(cx+d)+csin(ax+b)tan(cx+d)sec(cx+d)ddxsin(ax+b)cos(cx+d)=acos(ax+b)sec(cx+d)+csin(ax+b) tan(cx+d) sec(cx+d)

Q.40

Differentiate the function with respect to x.

cosx3. sin2(x5)

Ans.

Let y=cosx3. sin2(x5)Differentiating w.r.t. x, we getdydx=cosx3ddxsin2(x5)+sin2(x5)ddxcosx3  [By​ Product Rule.]      =cosx3ddx{sin(x5)}2+sin2(x5)×sinx3ddxx3      =cosx3×2sinx5ddxsin(x5)sin2(x5)sinx3×3x2      =2cosx3sinx5×cos(x5)ddxx53x2sin2(x5)sinx3      =2cosx3sinx5cos(x5)×5x43x2sin2(x5)sinx3      =10x4sinx5cos(x5)cosx33x2sinx3sin2(x5)ddxcosx3. sin2(x5)=10x4sinx5cos(x5)cosx33x2sinx3sin2(x5)

Q.41

Differentiate the function with respect to x. cos(x)

Ans.

Let y=cos(x)Differentiating w.r.t. x, we get  dydx=ddxcos(x)=sinxddxx=sinx×12x12=sinx2x

Q.42

Differentiate the function with respect to x. 2cot(x2)

Ans.

Let y=2cot(x2)Differentiating w.r.t. x, we getdydx=2ddxcot(x2)      =2×12{cot(x2)}12ddxcot(x2)  [By​ Chain Rule]      ={cot(x2)}12×cosec2(x2)ddxx2      ={cot(x2)}12×cosec2(x2)×2x      =2xcosec2(x2)cot(x2)      =2xsin2x2cosx2sinx2      =2xsinx2sinx2cosx2      =22xsinx22sinx2cosx2      =22xsinx2sin2x2ddx2cot(x2)=22xsinx2sin2x2

Q.43

Prove that the function f given by fx=x1,  x   is notdifferentiable at x=1.

Ans.

The given function is  f(x)=|x1|,  xRSince a function f is differentiable at a point c in its domain if bothlimx0f(c+h)f(c)handlimx0+f(c+h)f(c)h are finite and equal.Now, left hand limit at x=1,limh0f(1+h)f(1)h=limh0|1+h1||11|h=limh0|h|h=limh0hh(h<0|h|=h)=1Now, right hand limit at x=1,limh0+f(1+h)f(1)h=limh0+|1+h1||11|h=limh0|h|h=limh0hh(h>0|h|=h)=1Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x=1.

Q.44

Prove that the greatest integer function defined​ byfx=x,  0<x<3is not differentiable at​ x=1 and x=2.

Ans.

The given function f is  f(x)=[x],  0<x<3Since a function f is differentiable at a point c in its domain if bothlimx0f(c+h)f(c)handlimx0+f(c+h)f(c)h are finite and equal.Now, left hand limit at x=1,limh0f(1+h)f(1)h=limh0[1+h][1]h=limh001h=limh01h=And, left hand limit at x=1,limh0+f(1+h)f(1)h=limh0+[1+h][1]h=limh0+11h=limh0+0h=0Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1Checking the differentiability of the given function at x = 2,So, left hand limit at x=2,limh0f(2+h)f(2)h=limh0[2+h][2]h=limh012h=limh01h=And, left hand limit at x=2,limh0+f(2+h)f(2)h=limh0+[2+h][2]h=limh0+22h=limh0+0h=0Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2.

Q.45

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

Ans.

Given:  2x + 3y = sinxDifferentiating w.r.t. x, we getddx(2x + 3y)=ddxsinx2ddxx+3dydx=cosx        2+3dydx=cosx        dydx=cosx23

Q.46

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

Ans.

Given:  2x + 3y = sinyDifferentiating w.r.t. x, we getddx(2x + 3y)=ddxsiny2ddxx+3dydx=cosydydx        2+3dydx=cosydydx3dydxcosydydx=2  dydx(3cosy)=2    dydx=2(3cosy)    dydx=2(cosy3)

Q.47

Find dydx in the following: ax + by2 = cosy

Ans.

Given:  ax + by2 = cosyDifferentiating w.r.t. x, we getddx(ax + by2)=ddxcosyaddxx+2bydydx=sinydydx        a+2bydydx=sinydydx2bydydx+sinydydx=a  dydx(2by+siny)=a    dydx=a(2by+siny)    dydx=a(2by+siny)

Q.48

Find dydx in the following:xy+y2=tanx+y

Ans.

Given:  xy + y2=tanx+yDifferentiating w.r.t. x, we get  ddx(xy + y2)=ddx(tanx+y)[xdydx+yddxx]+ddxy2=sec2x+dydx[By​ Product Rule]            xdydx+y+2ydydx=sec2x+dydx        xdydx+2ydydxdydx=sec2xy              (x+2y1)dydx=sec2xy                  dydx=sec2xy(x+2y1)

Q.49

Find dydx in the following x3+ x2y + xy2+ y3= 81

Ans.

Given:  x3+ x2y + xy2+ y3= 81Differentiating w.r.t. x, we get                  ddx(x3+ x2y + xy2+ y3)=ddx81  ddxx3+ddxx2y+ddxxy2+ddxy3=03x2+(x2dydx+yddxx2)+(xddxy2+y2ddxx)+3y2dydx=0[By product rule]        3x2+x2dydx+y×2x+(2xydydx+y2×1)+3y2dydx=0                      3x2+x2dydx+2xy+2xydydx+y2+3y2dydx=0              (x2+2xy+3y2)dydx=3x22xyy2      dydx=(3x2+2xy+y2)(x2+2xy+3y2)

Q.50

Given:  x2+ xy + y2=100Differentiating w.r.t. x, we get      ddx(x2+ xy + y2)=ddx100  ddxx2+ddxxy+ddxy2=02x+(xdydx+yddxx)+2ydydx=0[By product rule]      2x+xdydx+y×1+2ydydx=0    dydx(x+2y)=2xydydx=2x+yx+2y

Ans.

Find dydx in the followingx2+ xy + y2= 100

Q.51

Find dydx in the followingx2+ xy + y2= 100

Ans.

Given:  x2+ xy + y2=100Differentiating w.r.t. x, we get      ddx(x2+ xy + y2)=ddx100  ddxx2+ddxxy+ddxy2=02x+(xdydx+yddxx)+2ydydx=0[By product rule]      2x+xdydx+y×1+2ydydx=0    dydx(x+2y)=2xydydx=2x+yx+2y

Q.52

sin2y+cosxy=K

Ans.

Given:  sin2y+cosxy=KDifferentiating w.r.t. x, we get                      ddxsin2y+cosxy=ddxK                      ddxsin2y+ddxcosxy=0              2sinyddxsinysinxyddxxy=0Using chain rule2sinycosydydxsinxyxdydx+yddxx=0By product rule          sin2ydydxxsinxydydxysinxy=0            dydxsin2yxsinxy=ysinxy        dydx=ysinxysin2yxsinxy

Q.53

Find dydx in the followingsin2x + cos2y = 1

Ans.

Given:  sin2x + cos2y=1Differentiating w.r.t. x, we get                            ddx(sin2x + cos2y)=ddx1                    ddxsin2x+ddxcos2y=02sinxddxsinx+2cosyddx(cosy)=0[Using chain rule]    2sinxcosx+2cosy(siny)dydx=0                sin2xsin2ydydx=0        sin2ydydx=sin2x    dydx=sin2xsin2y

Q.54

Find dydx in the followingy=sin1(2x1+x2)

Ans.

The given relationship is  y=sin1(2x1+x2)  siny=2x1+x2Differentiating both sides w.r.t. x, we get    ddxsiny=ddx(2x1+x2)      cosydydx=(1+x2)ddx2x2xddx(1+x2)(1+x2)2[By Quotient Rule]        1sin2ydydx=(1+x2)×22x(0+2x)(1+x2)2      1(2x1+x2)2dydx=2+2x24x2(1+x2)2          [Putting value of siny](1+x2)24x2(1+x2)2dydx=22x2(1+x2)2      (1x2)2(1+x2)2dydx=22x2(1+x2)2            (1x2)(1+x2)dydx=22x2(1+x2)2        dydx=22x2(1+x2)(1x2)      dydx=2(1x2)(1+x2)(1x2)      dydx=2(1+x2)

Q.55

Find dydx in the following:y=tan1(3xx31+3x2),13<x<13

Ans.

We​​ have y=tan1(3xx313x2)    Putting x=tanθ,​ we get          y=tan1(3tanθtan3θ13tan2θ)    =tan1(tan3θ)Since,13<x<13,thenx=tanθ  13<tanθ<13π6<θ<π6π2<3θ<π2    y=3θ[∵π2<3θ<π2]=3tan1x[∵x=tanθθ=tan1x]Differentiating both sides w.r.t. x, we get  dydx=3ddxtan1x=3×11+x2=31+x2ddxtan1(3xx313x2)=31+x2

Q.56

Find dydx in the following: y=cos1(1x21+x2),    0<x<1

Ans.

We​​ have y=cos1(1x21+x2),    0<x< 1    Putting x=tanθ,​ we get          y=cos1(1tan2θ1+tan2θ)    =cos1(cos2θ)Since,0<  x <  1,thenx=tanθ          0<tanθ<1    0<θ<π4    0<2θ<π2    y=2θ[∵0<2θ<π2]=2tan1x[∵x=tanθθ=tan1x]Differentiating both sides w.r.t. x, we getdydx=2ddxtan1x=2×11+x2=21+x2ddxcos1(1tan2θ1+tan2θ)=21+x2

Q.57

We​​ have y=cos1(1x21+x2),    0<x< 1    Putting x=tanθ,​ we get          y=cos1(1tan2θ1+tan2θ)    =cos1(cos2θ)Since,0<  x <  1,thenx=tanθ          0<tanθ<1    0<θ<π4    0<2θ<π2    y=2θ[∵0<2θ<π2]=2tan1x[∵x=tanθθ=tan1x]Differentiating both sides w.r.t. x, we getdydx=2ddxtan1x=2×11+x2=21+x2ddxcos1(1tan2θ1+tan2θ)=21+x2

Ans.

We​​ have y=sin1(1x21+x2),    0<x< 1    Putting x=tanθ,​ we get          y=sin1(1tan2θ1+tan2θ)    =sin1(cos2θ)    =sin1{sin(π22θ)}Since,0<  x <  1,then  x=tanθ          0<tanθ<1    0<θ<π4    0<2θ<π2      0<π22θ<π2    y=π22θ[∵0<π22θ<π2]=π22tan1x[∵x=tanθθ=tan1x]Differentiating both sides w.r.t. x, we getdydx=02ddxtan1x=2×11+x2=21+x2Thus,ddxsin1(1x21+x2)=21+x2

Q.58

Find dydx in the following:y=cos1(2x1+x2),    1<x<1

Ans.

We​​ have y=cos1(2x1+x2),    1<x<1    Putting x=tanθ,​ we get          y=cos1(2tanθ1+tan2θ)    =cos1(sin2θ)    =cos1{cos(π22θ)}Since,1<x<1,then  x=tanθ          1<tanθ<1  π4<θ<π4  π2<2θ<π2      π2>2θ>π2      π>π22θ>0      0<π22θ<π        y=π22θ=π22tan1x[∵x=tanθθ=tan1x]Differentiating both sides w.r.t. x, we getdydx=ddxπ22ddxtan1x=02×11+x2ddxcos1(2x1+x2)=21+x2.

Q.59

Find dydx in the followingy=sin1(2x1x2),    12<x<12

Ans.

We​​ have y=sin1(2x1x2),    12< x<12    Putting x=sinθ,​ we get          y=sin1(2sinθ1sin2θ)    =sin1(2sinθcosθ)    =sin1(sin2θ)Since,12< x<12,then  x=sinθ        12<sinθ<12      π4<θ<π4      π2<2θ<π2      y=2θ=2sin1x[x=sinθθ=sin1x]Differentiating w.r.t. x, we get  dydx=ddx(2sin1x)=21x2Thus,ddx{sin1(2x1x2)}=21x2

Q.60

Find dy/dx in the following

y = sec1(12x21),    0<x<12

Ans.

We have y=sec1(12x21),     0< x < 12    =cos1(2x21)Putting x=cosθ, ​we gety=cos1(2cos2θ1)    =cos1(cos2θ)Since,0<x <12        0<cosθ  <12[∵x=cosθ]        3π2<θ<7π4        0<θ<π4        0<2θ<π2y=2θ      =2cos1x[∵x=cosθθ=cos1x]Differentiating w.r.t. x, we get  dydx=2ddxcos1x=2(11x2)=21x2Thus,ddxsec1(12x21)=21x2.

Q.61

Differentiate the following w.r.t. x:exsinx

Ans.

Let y=exsinxDifferentiating both sides w.r.t. x, we get  dydx=ddxexsinx=sinxddxexexddxsinx(sinx)2[By Quotient Rule]=sinx.exexcosx(sinx)2=ex(sinxcosx)sin2x,  x,  nZ

Q.62

Differentiate the following w.r.t. x:esin1x

Ans.

Let y=esin1xDifferentiating both sides w.r.t. x, we get  dydx=ddxesin1x=esin1xddxsin1x[By Chain Rule]=esin1x×11x2=esin1x1x2,  x(1,1)

Q.63

Differentiate the following w.r.t. x: e x 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaWHebGaaCyAaiaahAgacaWHMbGaaCyzaiaahkhacaWHLbGaaCOBaiaahshacaWHPbGaaCyyaiaahshacaWHLbGaaeiiaiaahshacaWHObGaaCyzaiaabccacaWHMbGaaC4BaiaahYgacaWHSbGaaC4BaiaahEhacaWHPbGaaCOBaiaahEgacaqGGaGaaC4Daiaac6cacaWHYbGaaiOlaiaahshacaGGUaGaaeiiaiaahIhacaGG6aGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ieqacaWFLbWaaWbaaSqabeaacaWF4bWaaWbaaWqabeaacaWHZaaaaaaaaaa@64C9@

Ans.

Let y= e x 3 Differentiating both sides w.r.t. x, we get dy dx = d dx e x 3 = e x 3 d dx x 3 By Chain Rule = e x 3 ×3 x 2 =3 x 2 .e x 3 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabYeacaqGLbGaaeiDaiaabccacaqG5bGaeyypa0Jaaeyz amaaCaaaleqabaacbeGaa8hEamaaCaaameqabaGaa83maaaaaaaake aacaqGebGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOB aiaabshacaqGPbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgacaqGGa GaaeOyaiaab+gacaqG0bGaaeiAaiaabccacaqGZbGaaeyAaiaabsga caqGLbGaae4CaiaabccacaqG3bGaaeOlaiaabkhacaqGUaGaaeiDai aab6cacaqGGaGaaeiEaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGa ae4zaiaabwgacaqG0baabaGaaGPaVlaaykW7caaMc8+aaSaaaeaaca WGKbGaamyEaaqaaiaadsgacaWG4baaaiabg2da9maalaaabaGaamiz aaqaaiaadsgacaWG4baaaiaabwgadaahaaWcbeqaaiaa=Hhadaahaa adbeqaaiaa=ndaaaaaaaGcbaGaaCzcaiabg2da9iaabwgadaahaaWc beqaaiaa=Hhadaahaaadbeqaaiaa=ndaaaaaaOWaaSaaaeaacaWGKb aabaGaamizaiaadIhaaaGaamiEamaaCaaaleqabaGaaG4maaaakiaa xMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7daWadaqaaiaabkeacaqG5bGaaeiiaiaaboeacaqGObGa aeyyaiaabMgacaqGUbGaaeiiaiaabkfacaqG1bGaaeiBaiaabwgaai aawUfacaGLDbaaaeaacaWLjaGaeyypa0JaaeyzamaaCaaaleqabaGa a8hEamaaCaaameqabaGaa83maaaaaaGccqGHxdaTcaaIZaGaamiEam aaCaaaleqabaGaaGOmaaaaaOqaaiaaxMaacqGH9aqpcaaIZaGaamiE amaaCaaaleqabaGaaGOmaaaakiaab6cacaqGLbWaaWbaaSqabeaaca WF4bWaaWbaaWqabeaacaWFZaaaaaaaaaaa@AC6E@

Q.64

Differentiate the following w.r.t. x: sin(tan1ex)

Ans.

Let y=sin(tan1ex)Differentiating both sides w.r.t. x, we get  dydx=ddxsin(tan1ex)=cos(tan1ex)ddxtan1ex[By Chain Rule]=cos(tan1ex)×11+(ex)2×ddxex=cos(tan1ex)1+(ex)2×exddx(x)=excos(tan1ex)1+(ex)2×(1)=excos(tan1ex)1+e2x

Q.65

Differentiate the following w.r.t. x: log(cos ex)

Ans.

Let y=log(cosex)Differentiating both sides w.r.t. x, we get  dydx=ddxlog(cosex)=1cosex×ddxcosex[By Chain Rule]=1cosex×sinex×ddxex=sinexcosex×ex=exsinexcosex=extanex,  ex(2n+1)π2,nN

Q.66

Differentiate the following w.r.t. x:ex+ex2+...+ex5

Ans.

Let y=ex+ex2+...+ex5Differentiating both sides w.r.t. x, we get  dydx=ddx(ex+ex2+...+ex5)=ddxex+ddxex2+ddxex3+ddxex4+ddxex5=ex+ex2ddx(x2)+ex3ddx(x3)+ex4ddx(x4)+ex5ddx(x5)[Using chain rule]=ex+2xex2+3x2ex3+4x3ex4+5x4ex5

Q.67

Differentiate the following w.r.t. x:ex,x>0

Ans.

Let y=ex,x>0then,      y2=exDifferentiating both sides w.r.t. x, we get  ddxy2=ddx(ex)  2ydydx=exddx(x)[Using chain rule]      dydx=ex2y.12x    =ex4xex    dydx=ex4xex,  x>0

Q.68

Differentiate the following w.r.t. x:log(logx),x>1

Ans.

Let y=log(logx),x>1Differentiatingg both sides w.r.t. x, we get  ddxy=ddx{log(logx)}      dydx=1logxddx(logx)[Using chain rule]      dydx=1logx.1x      dydx=1xlogx,  x>1

Q.69

Differentiate the following w.r.t. x:cosxlogx,x>0

Ans.

Let y=cosxlogx,x>0Differentiating both sides w.r.t. x, we get  ddxy=ddx(cosxlogx)      dydx=logxddxcosxcosxddxlogx(logx)2[By  QuotientRule]      dydx=logx×sinxcosx×1x(logx)2      dydx=(xlogx.sinx+cosx)x(logx)2,  x>0

Q.70

Differentiate the following w.r.t. x: cos( logx+ e x ),x>0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaahseacaWHPbGaaCOzaiaahAgacaWHLbGaaCOCaiaahwgacaWHUbGaaCiDaiaahMgacaWHHbGaaCiDaiaahwgacaqGGaGaaCiDaiaahIgacaWHLbGaaeiiaiaahAgacaWHVbGaaCiBaiaahYgacaWHVbGaaC4DaiaahMgacaWHUbGaaC4zaiaabccacaWH3bGaaiOlaiaahkhacaGGUaGaaCiDaiaac6cacaqGGaGaaCiEaiaacQdaaeaacaWLjaGaaCzcaGqabiaa=ngacaWFVbGaa83CamaabmaabaGaa8hBaiaa=9gacaWFNbGaa8hEaiaa=TcacaWFLbWaaWbaaSqabeaacaWF4baaaaGccaGLOaGaayzkaaGaaiilaiaaykW7caWH4bGaeyOpa4JaaCimaaaaaa@6B35@

Ans.

Let y=cos( logx+ e x ),x>0 Differentiating both sides w.r.t. x, we get d dx y= d dx cos( logx+ e x ) dy dx =sin( logx+ e x ) d dx ( logx+ e x ) [ Usingchain rule ] dy dx =sin( logx+ e x )( 1 x + e x ) dy dx =( 1 x + e x ).sin( logx+ e x ),x>0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0EBA@

Q.71

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

Ans.

Let​       y=cosx.cos2x.cos3xTaking logarithm on both the sides, we get      logy=log(cosx.cos2x.cos3x)      =logcosx+logcos2x+logcos3xDifferentiating both sides with respect to x, we getddxlogy=ddxlogcosx+ddxlogcos2x+ddxlogcos3x    1ydydx  =1cosxddxcosx+1cos2xddxcos2x+1cos3xddxcos3x    1ydydx  =1cosx×sinx+1cos2x×sin2xddx2x+1cos3x×sin3x×ddx3x    1ydydx  =sinxcosxsin2xcos2x×2sin3xcos3x×3      dydx  =y(tanx+2tan2x+3tan3x)      dydx  =cosx.cos2x.cos3x(tanx+2tan2x+3tan3x)

Q.72

Differentiate the function given with respect to x.

(x1)(x2)(x3)(x4)(x5)

Ans.

Let      y=(x1)(x2)(x3)(x4)(x5)Taking logarithm on both the sides, we getlogy=log(x1)(x2)(x3)(x4)(x5)    logy=12log{(x1)(x2)(x3)(x4)(x5)}    logy=12{log(x1)+log(x2)log(x3)log(x4)log(x5)}Differentiating both sides with respect to x, we getddx  logy=12ddx{log(x1)+log(x2)log(x3)log(x4)log(x5)}      1ydydx=12(1x1ddx(x1)+1x2ddx(x2)1x3ddx(x3)1x4ddx(x4)1x5ddx(x5))  [By chain rule]      1ydydx=12(1x1×1+1x2×11x3×11x4×11x5×1)          dydx=12y(1x1+1x21x31x41x5)          dydx=12(x1)(x2)(x3)(x4)(x5)(1x1+1x21x31x41x5)

Q.73

Differentiate the function given with respect to x.

(logx)cosx

Ans.

Let​       y=(logx)cosxTaking logarithm on both the sides, we get      logy=cosx.log(logx)Differentiating both sides with respect to x, we getddxlogy=ddx{cosx.log(logx)}      1ydydx=cosxddxlog(logx)+log(logx)ddxcosx[By product rule]      1ydydx=cosx.1logxddxlogx+log(logx).(sinx)[By Chain rule]        dydx=y{cosxlogx×1xsinx.log(logx)}      dydx=(logx)cosx{cosxxlogxsinx.log(logx)}

Q.74

Differentiate the function given with respect to x.

xx2sinx

Ans.

Let​       y=xx2sinx        =y1+y2So,    y1=xxTaking logarithm on both the sides, we get      logy1=log(xx)=xlogxDifferentiating both sides with respect to x, we get1y1dy1dx=xddxlogx+logxddxx    =x.1x+logx.1    =1+logx    dy1dx=y1(1+logx)    =xx(1+logx)Now,  y2=2sinxTaking logarithm on both the sides, we get      logy2=log(2sinx)=sinx.log2Differentiating both sides with respect to x, we get1y2dy2dx=log2ddxsinx    =log2.cosx    dy1dx=y2(log2.cosx)    =2sinxcosx.log2Thus,dydx=dy1dxdy2dx=xx(1+logx)2sinxcosx.log2

Q.75

Differentiate the function given with respect to x.

(x+3)2.(x+4)3.(x+5)4

Ans.

Let​       y=(x+3)2.(x+4)3.(x+5)4Taking logarithm on both the sides, we get      logy=log{(x+3)2.(x+4)3.(x+5)4}      =2log(x+3)+3log(x+4)+4log(x+5)Differentiating both sides with respect to x, we getddxlogy=2ddxlog(x+3)+3ddxlog(x+4)+4ddxlog(x+5)      1ydydx=2×1x+3ddx(x+3)+3×1x+4ddx(x+4)+4×1x+5ddx(x+5)      =2x+3×1+3x+4×1+4x+5×1        dydx=y(2x+3+3x+4+4x+5)      =(x+3)2.(x+4)3.(x+5)4(2x+3+3x+4+4x+5)      =(x+3)2.(x+4)3.(x+5)4(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{2(x2+9x+20)+3(x2+8x+15)+4(x2+7x+12)}      =(x+3).(x+4)2.(x+5)3{2x2+18x+40+3x2+24x+45+4x2+28x+48}      dydx=(x+3).(x+4)2.(x+5)3(9x2+70x+133)

Q.76

Differentiate the function given with respect to x.

(x+1x)x+x(1+1x)

Ans.

Let​       y=(x+1x)x+x(1+1x)        =y1+y2So,    y1=(x+1x)xTaking logarithm on both the sides, we get      logy1=xlog(x+1x)Differentiating both sides with respect to x, we getddxlogy1=ddx{xlog(x+1x)}    1y1dy1dx=xddxlog(x+1x)+log(x+1x)ddxx    1y1dy1dx=x.1(x+1x)ddx(x+1x)+log(x+1x)×1        dy1dx=y1{x2x2+1×(11x2)+log(x+1x)}        dy1dx=(x+1x)x{x2x2+1×(x21x2)+log(x+1x)}        dy1dx=(x+1x)x{x21x2+1+log(x+1x)}Now,    y2=x(1+1x)Taking logarithm on both the sides, we get      logy2=(1+1x)logxDifferentiating both sides with respect to x, we getddxlogy2=ddx{(1+1x)logx}    1y2dy2dx=(1+1x)ddxlogx+logxddx(1+1x)      [By​ product rule]        dy2dx=y2{(1+1x)1x+logx.(01x2)}        dy2dx=x(1+1x){1x+1x21x2logx}=x(1+1x){x+1logxx2}Thus,        dydx=dy1dx+dy2dx                =(x+1x)x{x21x2+1+log(x+1x)}+x(1+1x){x+1logxx2}

Q.77

Differentiate the function given with respect to x.

(logx)x+xlogx

Ans.

Let​  y=(logx)x+xlogx         =y1+y2So,    y1=(logx)xTaking logarithm on both the sides, we get      logy1=x.log(logx)Differentiating both sides with respect to x, we getddxlogy1=ddx{x.log(logx)}    1y1dy1dx=xddxlog(logx)+log(logx)ddxx[By Product Rule]    1y1dy1dx=x.1logxddxlogx+log(logx)×1        dy1dx=y1{x.1logx1x+log(logx)}        dy1dx=(logx)x{1logx+log(logx)}        dy1dx=(logx)x-1{1+logx.log(logx)}Now,  y2=xlogxTaking logarithm on both the sides, we get      logy2=logx.logx        =(logx)2Differentiating both sides with respect to x, we getddxlogy2=ddx(logx)21y2dy2dx=2logxddxlogx[By Chain Rule]    =2logx×1x    dy2dx=2y2xlogx    =2xlogxxlogx    =2xlogx-1logxHence,        dydx=dy1dx+dy2dx    =(logx)x-1{1+logx.log(logx)}+2xlogx-1logx

Q.78

Differentiate the function given with respect to x.

(sinx)x+sin1x

Ans.

Let​  y=(sinx)x+sin1x       =y1+y2So,    y1=(sinx)xTaking logarithm on both the sides, we get      logy1=x.log(sinx)Differentiating both sides with respect to x, we getddxlogy1=ddx{x.log(sinx)}    1y1dy1dx=xddxlog(sinx)+log(sinx)ddxx[By Product Rule]    1y1dy1dx=x.1sinxddxsinx+log(sinx)×1        dy1dx=y1{x.1sinxcosx+log(sinx)}        dy1dx=(sinx)x{xcosxsinx+log(sinx)}        dy1dx=(sinx)x{xcotx+log(sinx)}Now,  y2=sin1xDifferentiating both sides with respect to x, we getdy2dx=ddxsin1xdy2dx=11(x)2ddxx[By Chain Rule]    =11x×12x    dy2dx=12xx2Hence,        dydx=dy1dx+dy2dx    =(sinx)x{xcotx+log(sinx)}+12xx2

Q.79

Differentiate the function given with respect to x.

xsinx+(sinx)cosx

Ans.

Let​  y=xsinx+(sinx)cosx       =y1+y2So,    y1=xsinxTaking logarithm on both the sides, we get      logy1=sinx.logxDifferentiating both sides with respect to x, we getddxlogy1=ddx{sinx.logx}    1y1dy1dx=sinxddxlogx+logxddxsinx[By Product Rule]    1y1dy1dx=sinx.1x+logx×cosx        dy1dx=y1{sinx.1x+logx×cosx}        dy1dx=xsinx{sinxx+cosx.logx}Now,  y2=(sinx)cosxTaking logarithm on both the sides, we get      logy1=cosx.logsinxDifferentiating both sides with respect to x, we getddxlogy1=ddx(cosx.logsinx)  1y2dy2dx=cosxddxlogsinx+logsinxddxcosx[By Product Rule]  1y2dy2dx=cosx.1sinxddxsinx+logsinx×sinx      dy2dx=(sinx)cosx(cosx.1sinx.cosxsinx.logsinx)      =(sinx)cosx(cotx.cosxsinx.logsinx)      dydx=dy1dx+dy2dx      =xsinx(sinxx+cosx.logx)+(sinx)cosx(cosx.cotxsinx.logsinx)

Q.80

Differentiate the function given with respect to x.

xxcosx+x2+1x21

Ans.

Lety=xxcosx+x2+1x21         =y1+y2So,    y1=xxcosxTaking logarithm on both the sides, we getlogy1=xcosx.logxDifferentiating both sides with respect to x, we getddxlogy1=ddx{xcosx.logx}    1y1dy1dx=xcosxddxlogx+logxddx(xcosx)        [By Product Rule]    1y1dy1dx=xcosx.1x+logx(xddxcosx+cosxddxx)[By Product Rule]        dy1dx=y1{cosx+logx(xsinx+cosx)}        dy1dx=xxcosx{cosxxsinxlogx+logx.cosx}        dy1dx=xxcosx{cosx(1+logx)xsinxlogx}Now,                y2=x2+1x21Taking logarithm on both the sides, we get            logy2=log(x2+1)log(x21)Differentiating w.r.t. x, we get          1y2dy2dx=1x2+1ddx(x2+1)1x21ddx(x21)[By chain rule]              dy2dx=y2(1x2+1×2x1x21×2x)            =2x(x2+1x21)(1x2+11x21)    =2x(x2+1x21)((x21)(x2+1)(x2+1)((x21)))    =2x(x2+1x21){x21x21(x2+1)(x21)}    =2x(1x21){2(x21)}  dy2dx=4x(x21)2Thus,     dydx=dy1dx+dy2dx=xxcosx{cosx(1+logx)xsinxlogx}4x(x21)2

Q.81

Differentiate the function given with respect to x. (xcosx)x+(xsinx)1x

Ans.

Let​       y=(xcosx)x+(xsinx)1x        =y1+y2So,    y1=(xcosx)xTaking logarithm on both the sides, we get      logy1=xlog(xcosx)        =xlogx+xlogcosxDifferentiating both sides with respect to x, we getddxlogy1=ddx(x.logx+x.logcosx)    1y1dy1dx=ddx(x.logx)+ddx(x.logcosx)    1y1dy1dx=xddxlogx+logxddxx+xddxlogcosx+logcosxddxx[By Product Rule]        dy1dx=y1(x.1x+logx+x1cosx.ddxcosx+logcosx)        dy1dx=(xcosx)x(1+logx+x1cosx×sinx+logcosx)      =(xcosx)x(1+logxxsinxcosx+logcosx)        dy1dx=(xcosx)x(1+logxxtanx+logcosx)Now,                y2=(xsinx)1xTaking logarithm on both the sides, we get      logy2=1xlog(xsinx)Differentiating w.r.t. x, we get        1y2dy2dx=1xddxlog(xsinx)+log(xsinx)ddx.1x[By product rule]            dy2dx=y2{1x(1xsinx)ddx(xsinx)log(xsinx).1x2}[By chain rule]            dy2dx=(xsinx)1x{1x(1xsinx)(xddxsinx+sinxddxx)log(xsinx).1x2}            dy2dx=(xsinx)1x{1x(1xsinx)(x.cosx+sinx)log(xsinx).1x2}            dy2dx=(xsinx)1x{(x.cosx+sinxx2sinx)log(xsinx).1x2}            dydx=dy1dx+dy2dx    dydx=(xcosx)x(1+logxxtanx+logcosx)+(xsinx)1x{(x.cosx+sinxx2sinx)log(xsinx).1x2}

Q.82

Find dydx of the function xy+yx=

Ans.

The given function is      xy+yx=1Let u=xy and v=yxThen, function becomes    u+v=1Differentiating w.r.t. x, we get        dudx+dvdx=0...(i)Since,    u=xyTaking log both sides, we getlogu=logxylogu=ylogxDifferentiating both sides w.r.t. x, we get  ddxlogu=ddxylogx1ududx=yddxlogx+logxdydx=y×1x+logxdydx  dudx=u(yx+logxdydx)  dudx=xy(yx+logxdydx)  ...(ii)And        v=yxTaking log both sides, we getlogv=logyxlogv=xlogyDifferentiating both sides w.r.t. x, we get  ddxlogv=ddxxlogy        1vdvdx=xddxlogy+logyddxx        =x×1ydydx+logy(1)  dvdx=v(xydydx+logy)  dvdx=yx(xydydx+logy)...(iii)From equatin(i),(ii) and (iii), we havexy(yx+logxdydx)+yx(xydydx+logy)=0xy1y+xylogxdydx+yx1xdydx+yxlogy=0(xylogx+yx1x)dydx=xy1yyxlogy    dydx=yxy1+yxlogyxylogx+xyx1

Q.83

Find dydx of the function : yx=xy

Ans.

The given function is yx=xyTaking logarithm both sides, we get      logyx=logxy        xlogy=ylogxDifferentiating both sides, w.r.t. x, we getxddxlogy+logyddxx=yddxlogx+logxddxy[ByProduct Rule]  xydydx+logy×1=yx+logxdydx  xydydxlogxdydx=yxlogy    (xylogx)dydx=yxlogy      dydx=(yxlogy)(xylogx)  =y(yxlogy)x(xylogx)

Q.84

Find dydx of the function : cosxy = cosyx

Ans.

The given function is (cosx)y=(cosy)xTaking logarithm both sides, we get      log(cosx)y=log(cosy)x            ylogcosx=xlogcosyDifferentiating both sides, w.r.t. x, we get      yddxlogcosx+logcosxddxy=xddxlogcosy+logcosyddxx[ByProduct Rule]y×1cosxddxcosx+logcosxdydx=x×1cosyddxcosy+logcosy      ycosx×sinx+logcosxdydx=xcosy×sinydydx+logcosy              ysinxcosx+logcosxdydx=xsinycosydydx+logcosy              logcosxdydx+xsinycosydydx=logcosy+ysinxcosx                (logcosx+xsinycosy)dydx=logcosy+ysinxcosx      dydx=(logcosy+ysinxcosx)(logcosx+xsinycosy)  =(logcosy+ytanx)(logcosx+xtany)

Q.85

Find dydx of the function :xy = exy

Ans.

The given function is xy=e(xy)Taking logarithm both sides, we get      logxy=loge(xy)                logx+logy=(xy)loge                logx+logy=(xy)×1∵[loge=1]                logx+logy=(xy)Differentiating both sides, w.r.t. x, we get        ddxlogx+ddxlogy=ddxxdydx[ByProduct Rule]  1x+1ydydx=1dydx  (1+1y)dydx=11x  dydx=11x(1+1y)=(x1x)(y+1y)  dydx=y(x1)x(y+1)

Q.86

Find the derivative of the function given by  fx=1+x1+x21+x41+x8and hence find f 1.

Ans.

The given relationship is  f(x)=(1+x)(1+x2)(1+x4)(1+x8)Taking logarithm on both the sides, we getlogf(x)=log(1+x)+log(1+x2)+log(1+x4)+log(1+x8)Differentiating both sides with respect to x, we get1f(x)ddxf(x)=11+xddx(1+x)+11+x2ddx(1+x2)+11+x4ddx(1+x4)+11+x8ddx(1+x8)      =11+x(1)+11+x2(0+2x)+11+x4(0+4x3)+11+x8(0+8x7)ddxf(x)=f(