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:

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 

Provethatthefunctionfx=5x3iscontinuousatx=0,atx=3andatx=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

Examinethefollowingfunctionsforcontinuity.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

Provethatthefunctionf x = x n iscontinuousatx = n, wherenisapositiveinteger.

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

Isthefunctionfdefinedbyfx=x,  if  x15,  if  x>1continuousatx=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

Findtherelationshipbetweenaandbsothatthefunctionfdefinedbyfx=ax+1,  if  x3bx+,  if  x>3tinuousatx=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

Forwhatvalueofλisthefunctionfdefinedbyfx=λx22x,  if  x04x+,  if  x>0tinuousatx=0?Whataboutcontinuityatx=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

Findthepointsofdiscontinuityoff,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

Determineiffdefinedbyfx=x2sinx,  if  x0,  if  x=0isacontinuousfunction?

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

Examinethecontinuityoff,wherefisdefinedbyfx=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

Findthevaluesofaandbsuchthatthefunctiondefinedbyfx=,  if  xax+b,  if<x<,if  xisacontinuousfunction.

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. 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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

Provethatthefunctionfgivenbyfx=x1,  xisnotdifferentiableatx=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

Provethatthegreatestintegerfunctiondefinedbyfx=x,  0<x<3isnotdifferentiableatx=1andx=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

Finddydxofthefunctionxy+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

Finddydxofthefunction: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

Finddydxofthefunction: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

Finddydxofthefunction: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

Findthederivativeofthefunctiongivenby  fx=1+x1+x21+x41+x8andhencefindf1.

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(x)(11+x+2x1+x2+4x31+x4+8x71+x8)f(x)=(1+x)(1+x2)(1+x4)(1+x8)(11+x+2x1+x2+4x31+x4+8x71+x8)Putting x = 1, we get  f(1)=(1+1)(1+12)(1+14)(1+18)(11+1+2(1)1+12+4(1)31+14+8(1)71+18)    =2×2×2×2(12+22+42+82)    =16(152)f(1)=120

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.

Let  y=(x2 5x + 8)(x3+ 7x + 9)(i) Differentiating w.r.t. x, by using product rule, we get    dydx=ddx{(x2 5x + 8)(x3+ 7x + 9)}=(x2 5x + 8)ddx(x3+ 7x + 9)+(x3+ 7x + 9)ddx(x2 5x + 8)=(x2 5x + 8)(3x2+ 7)+(x3+7x+9)(2x5)=3x4+7x215x335x+24x2+56+2x45x3+14x235x+18x45    dydx=5x420x3+45x252x+11(ii)y=(x25x + 8)(x3+ 7x + 9)=x5+7x3+9x25x435x245x+8x3+56x+72=x55x4+15x326x2+11x+72Differentiating w.r.t. x,we get  dydx=ddx(x55x4+15x326x2+11x+72)=5x420x3+45x252x+11(iii) y=(x25x + 8)(x3+ 7x + 9)Taking logarithm on both the sides, we getlogy=log{(x25x + 8)(x3+ 7x + 9)}=log(x25x + 8)+log(x3+ 7x + 9)Differentiating w.r.t. x,we getddxlogy=ddxlog(x25x + 8)+ddxlog(x3+ 7x + 9)      =1(x25x + 8)ddx(x25x + 8)+1(x3+ 7x + 9)ddx(x3+ 7x + 9)1ydydx=1(x25x + 8)(2x5)+1(x3+ 7x + 9)(3x2+7)  dydx=y((2x5)(x3+ 7x + 9)+(3x2+7)(x25x + 8)(x25x + 8)(x3+ 7x + 9))  dydx=y((2x5)(x3+ 7x + 9)+(3x2+7)(x25x + 8)y)[∵Lety=(x25x + 8)(x3+ 7x + 9)  dydx=2x4+14x2+18x5x335x45+3x415x3+24x2+7x235x+56  dydx=5x420x3+45x252x+11From the above three observations, it can be say that all the results are same.

Q.88

Ifu,vandwarefunctionsofx,thenshowthatddxu, v, w=dudxv.w+udvdx.w+u.v.dwdxintwowaysfirstbyrepeatedapplicationofproductrule,secondbylogarithmicdifferentiation.

Ans.

Let  y=u.v.w=u.(v.w)By applying product rule, we get    dydx=ddx{u.(v.w)}=uddx(v.w)+(v.w)ddxu=u{vddxw+wddxv}+(v.w)ddxu=u.vddxw+u.wddxv+(v.w)ddxu  dydx=dudx.v.w+dvdx.u.w+dwdx.u.vThus, this is the differentiation by product rule.By taking logarithm on both sides of the equationy=u.v.w,we getlogy=logu+logv+logwDifferentiating both sides with respect to x, we get1ydydx=1ududx+1vdvdx+1wdwdx  dydx=y(1ududx+1vdvdx+1wdwdx)=u.v.w(1ududx+1vdvdx+1wdwdx)  dydx=dudx.v.w+dvdx.u.w+dwdx.u.vThus, this is the differentiation by logarithmic differentiation.

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.

The given equations are  x=2at2, y=at4Differentiating w.r.t. t, we getdxdt=ddt2at2=4atanddydt=ddtat4=4at3dydx=(dydt)(dxdt)=4at34at=t2

Q.90

Ifxandyareconnectedparametricallybythegivenequations, withouteliminatingtheparameter,finddydx.x=a cosθ, y = b cosθ

Ans.

The given equations are  x=acosθ, y=bcosθDifferentiating w.r.t. θ, we getdx=dacosθ=asinθanddy=dbcosθ=bsinθdydx=(dy)(dx)=bsinθasinθ=ba

Q.91

Ifxandyareconnectedparametricallybythegivenequations, withouteliminatingtheparameter,finddydx.x=sint, y = cos2t

Ans.

The given equations are  x=sint, y=cos2tDifferentiating w.r.t. t, we getdxdt=ddtsint=costanddydt=ddtcos2t      =2sin2t      =4sintcostdydx=(dy)(dx)=4sintcostcost=4sint

Q.92

Ifxandyareconnectedparametricallybythegivenequations, withouteliminatingtheparameter,finddydx.x=4t, y = 4t

Ans.

The given equations are  x=4t, y=4tDifferentiating w.r.t. t, we getdxdt=ddt4t=4anddydt=ddt4t=4t2dydx=(dydt)(dxdt)=4t24dydx=1t2

Q.93

Ifxandyareconnectedparametricallybythegivenequations, withouteliminatingtheparameter,finddydx.x=cosθcos2θ y = sinθsin2θ

Ans.

The given equations arex=cosθcos2θ, y=sinθsin2θ Differentiating w.r.t. θ, we get dx dθ = d dθ ( cosθcos2θ ) =sinθ+2sin2θ and dy dθ = d dθ ( sinθsin2θ ) =cosθ2cos2θ dy dx = ( dy dθ ) ( dx dθ ) = cosθ2cos2θ 2sin2θsinθ MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@06BB@

Q.94

Ifxandyareconnectedparametricallybythegivenequations, withouteliminatingtheparameter,finddydx.x = aθsinθ, y = a1+cosθ

Ans.

The given equations are  x=a(θsinθ), y=a(1+cosθ)Differentiating w.r.t. θ, we get  dx=da(θsinθ)=a(1cosθ)and  dy=da(1+cosθ)=a(0sinθ)=asinθdydx=(dy)(dx)=asinθa(1cosθ)=2sinθ2cosθ22sin2θ2=cotθ2

Q.95

Ifxandyareconnectedparametricallybythegivenequations, withouteliminatingtheparameter,finddydx.x=sin3tcos2t y = cos3tcos2t

Ans.

The given equations are  x=sin3tcos2t, y=cos3tcos2tDifferentiating w.r.t. t, we get  dxdt=ddtsin3tcos2t=cos2tddtsin3tsin3tddtcos2t12cos2t2=cos2t3sin2tddtsintsin3t×12cos2t12ddtcos2tcos2t=cos2t3sin2t×costsin3t×12cos2t12×2sin2tcos2t=3cos2tsin2tcost+sin3tsin2tcos2tcos2tanddydt=ddtcos3tcos2t=cos2tddtcos3tcos3tddtcos2t12cos2t2=cos2t3cos2tddtcostcos3t×12cos2t12ddtcos2tcos2t=cos2t3cos2t×sintcos3t×12cos2t12×2sin2tcos2t=3cos2tcos2tsint+cos3tsin2tcos2tcos2tdydx=dydtdxdt=3cos2tcos2tsint+cos3tsin2tcos2tcos2t3cos2tsin2tcost+sin3tsin2tcos2tcos2tdydx=3cos2tcos2tsint+cos3t×2sintcost3cos2tsin2tcost+sin3t×2sintcost∵sin2t=2sintcost=sintcost3cos2tcost+2cos3tsintcost3cos2tsint+2sin3t=32cos2t1cost+2cos3t312sin2tsint+2sin3t=6cos3t+3cost+2cos3t3sint6sin3t+2sin3t=4cos3t+3cost3sint4sin3t=cos3tsin3t

Q.96

Ifxandyareconnectedparametricallybythegivenequations, withouteliminatingtheparameter,finddydx.x =a cos t+log tant2, y = a sin t

Ans.

The given equations are  x=a(cost + logtant2), y=asintDifferentiating w.r.t. t, we get  dxdt=ddta(cost + logtant2)=a(ddtcost + ddtlogtant2)=a(sint + 1tan(t2)ddttan(t2))[Bychain rule]=a(sint + 1tan(t2).sec2(t2)ddtt2)=a(sint + 1tan(t2).sec2(t2)×12)=a(sint + 121sin(t2)cos(t2))=a(sint + 1sin2(t2))[∵sin2θ=2sinθcosθ]=a(sint + 1sint)=a(1sin2tsint)=acos2tsint anddydt=ddt asint=acostdydx=(dydt)(dxdt)=acost(acos2tsint)=sintcostdydx=tant

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


Ans.

The given equations are  x=asecθ, y=b tanθDifferentiating w.r.t. θ, we get  dx=dasecθ=asecθtanθand  dy=db tanθ=bsec2θdydx=(dy)(dx)=bsec2θasecθtanθ=ba.secθ.cotθ=ba.1cosθ.cosθsinθ=ba×1sinθ=bacosecθ

Q.98

Ifxandyareconnectedparametricallybythebelowequations,withouteliminatingtheparameter,finddydx.x= acosθ+θsinθ,y=asinθθcosθ

Ans.

The given equations are  x=a(cosθ+θsinθ), y=a(sinθθcosθ)Differentiating w.r.t. θ, we get  dx=d a(cosθ+θsinθ)=a(sinθ+θdsinθ+sinθdθ)=a(sinθ+θcosθ+sinθ×1)=aθcosθand  dy=da(sinθθcosθ)=a(cosθθdcosθcosθdθ)=a(cosθ+θsinθcosθ×1)=aθsinθdydx=(dy)(dx)=aθsinθaθcosθ=tanθ

Q.99

asin1t, y=acos1t,showthatdydx=yx

Ans.

The given equations are  x=asin1t=a12sin1t,    y=acos1t=a12cos1tTaking logarithm both sides, we getlogx=loga12sin1t=12sin1t.logaandlogy=loga12cos1t=12cos1t.logaDifferentiating w.r.t. t, we getddtlogx=ddt12sin1t.loga1xdxdt=12logaddtsin1t=12loga×11t2  dxdt=x(loga21t2)andddtlogy=ddt12cos1t.loga1ydydt=12logaddtcos1t=12loga×11t2  dydt=y(loga21t2)dydx=(dydt)(dxdt)=y(loga21t2)x(loga21t2)dydx=yxHence,  dydx=yx

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

Ans.

Lety=x2+3x+2Differentiating both sides w.r.t. x, we get  dydx=ddx(x2+3x+2)=2x+3Differentiatingagain w.r.t. x, we getd2ydx2=ddx(2x+3)  =2Thus, second derivative of given function is 2.

Q.101

Findthesecondorderderivativeof‹thefunction:x20

Ans.

Lety=x20Differentiating both sides w.r.t. x, we get  dydx=ddxx20=20x19Differentiatingagain w.r.t. x, we getd2ydx2=ddx(20x19)  =20×19x18  =380x18Thus, second derivative of given function is 380x18.

Q.102

Findthesecondorderderivativeofthefunction:x.cosx

Ans.

Lety=xcosxDifferentiating both sides w.r.t. x, we get  dydx=ddx(xcosx)=xddx cosx+cosxddxx[By product rule]=x(sinx)+cosx×1=xsinx+cosxDifferentiatingagain w.r.t. x, we getd2ydx2=ddx(xsinx+cosx)  =(xddxsinx+sinxddxx)+ddxcosx  =(xcosx+sinx×1)sinx  =xcosxsinxsinx  =xcosx2sinxThus, second derivative of given function is (xcosx+2sinx).

Q.103

Findthesecondorderderivative‹of‹thefunction:logx

Ans.

Lety=logxDifferentiating both sides w.r.t. x, we get  dydx=ddxlogx=1x=x1Differentiatingagain w.r.t. x, we getd2ydx2=ddx(x1)   =1x2   =x2Thus, second derivative of given function is x2.

Q.104

Findthesecondorderderivativeofthefunction:x3logx

Ans.

Lety=x3logxDifferentiating both sides w.r.t. x, we get  dydx=ddx(x3logx)=x3ddxlogx+logxddxx3[By product rule]=x3×1x+logx×3x2=x2+3x2logxDifferentiatingagain w.r.t. x, we getd2ydx2=ddx(x2+3x2logx)  =ddxx2+3ddx(x2logx)  =2x+3(x2ddxlogx+logxddxx2)  =2x+3(x2×1x+logx.2x)  =2x+3x+6xlogx  =5x+6xlogxThus, second derivative of given function is x (5+6logx).

Q.105

Findthesecondorderderivativeofthefunction:exsin5x

Ans.

Lety=exsin5xDifferentiating both sides w.r.t. x, we get  dydx=ddxexsin5x=exddxsin5x+sin5xddxexBy product rule=ex5cos5x+sin5x.ex=ex5cos5x+sin5xDifferentiating again w.r.t. x, we getd2ydx2=ddxex5cos5x+sin5x  =exddx5cos5x+sin5x+5cos5x+sin5xddxexBy product rule  =ex25sin5x+5cos5x+5cos5x+sin5xex  =ex25sin5x+5cos5x+5cos5x+sin5x  =2ex12sin5x+5cos5x

Q.106

Findthesecondorderderivativeofthefunction:e6xcos3x

Ans.

Lety=e6xcos3xDifferentiating both sides w.r.t. x, we get  dydx=ddx(e6xcos3x)=e6xddxcos3x+cos3xddxe6x[By product rule]=e6x(3sin3x)+cos3x.6e6x=e6x(6cos3x3sin3x)Differentiatingagain w.r.t. x, we getd2ydx2=ddx{e6x(6cos3x3sin3x)}  =e6xddx(6cos3x3sin3x)+(6cos3x3sin3x)ddxe6x[By product rule]  =ex(18sin3x9cos3x)+(6cos3x3sin3x)6e6x  =e6x(18sin3x9cos3x+36cos3x18sin3x)  =e6x(36sin3x+27cos3x)  =9e6x(4sin3x+3cos3x)Thus, second derivative of given function is 9e6x(3cos3x4sin3x).

Q.107

Findthesecondorderderivativeofthefunction:tan1x

Ans.

Lety=tan1xDifferentiating both sides w.r.t. x, we get  dydx=ddxtan1x=11+x2Differentiatingagain w.r.t. x, we getd2ydx2=ddx(11+x2)  =ddx(1+x2)1  =(1+x2)2ddx(1+x2)=(1+x2)2(0+2x)=2x(1+x2)2Thus, second derivative of given function is 2x(1+x2)2.

Q.108

Find the scond orderderivative of the function:log(logx)

Ans.

Lety=log(logx)Differentiating both sides w.r.t. x, we get  dydx=ddxlog(logx)=1logxddxlogx[By  chain  rule]=1logx(1x)=1xlogxDifferentiatingagain w.r.t. x, we getd2ydx2=ddx(1xlogx)  =ddx(xlogx)1  =(xlogx)2ddx(xlogx)[By  chain  rule]  =1(xlogx)2(xddxlogx+logxddxx)  =1(xlogx)2(x×1x+logx×1)  =1(xlogx)2(1+logx)  =(1+logx)(xlogx)2Thus, second derivative of given function is (1+logx)(xlogx)2.

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

Ans.

Lety=sin(logx)Differentiating both sides w.r.t. x, we get  dydx=ddxsin(logx)=cos(logx)ddxlogx[By  chain  rule]=cos(logx)(1x)=cos(logx)xDifferentiatingagain w.r.t. x, we getd2ydx2=ddx(cos(logx)x)  =xddxcos(logx)cos(logx)ddxxx2  =x×sin(logx)ddxlogxcos(logx)×1x2[By  chain  rule]  =xsin(logx)×1xcos(logx)x2  ={sin(logx)+cos(logx)}x2Thus, second derivative of given function is {sin(logx)+cos(logx)}x2.

Q.110

If y=5cosx3sinx, prove that d2ydx2+y=0

Ans.

Given:    y=5cosx3sinxDifferentiating w.r.t. x, we get  dydx=ddx(5cosx3sinx)=5ddxcosx3ddxsinx=5sinx3cosxAgain differentiating w.r.t. x, we get  d2ydx2=ddx(5sinx+3cosx)  =(5ddxsinx+3ddxcosx)  =(5cosx3sinx)  =yd2ydx2+y=0Hence Proved.

Q.111

If y=cos1x. Findd2ydx2 in term so fy alone.

Ans.

Given:    y=cos1x      cosy=xDifferentiating w.r.t. y, we getddycosy=ddyx      dxdy=siny      dydx=1siny      dydx=cosecy  ...(i)Again differentiating w.r.t. x, we get  d2ydx2=ddx(cosecy)  =cosecycotydydx  =cosecycoty×cosecy  =cosec2ycoty [From equation (i)]

Q.112

Ify=3cos(logx)+ 4sin(logx), show that x2y2+xy1+y=0

Ans.

Given:    y=3cos(logx) + 4 sin(logx)Differentiating w.r.t. x, we getdydx(=y1)=ddx{3cos(logx) +4 sin(logx)}=3ddxcos(logx) + 4ddxsin(logx)=3sin(logx)ddxlogx+4cos(logx)ddxlogx[By chain rule]=3sin(logx)1x+4cos(logx)1x=3sin(logx)+4cos(logx)xAgain differentiating w.r.t. x, we get  d2ydx2(y2)=ddx(4cos(logx)3sin(logx)x)=xddx{4cos(logx)3sin(logx)}{4cos(logx)3sin(logx)}ddxxx2=x{4sin(logx)ddxlogx3cos(logx)ddxlogx}{4cos(logx)3sin(logx)}×1x2=x{4sin(logx)1x3cos(logx)1x}{4cos(logx)3sin(logx)}x2=4sin(logx)3cos(logx)4cos(logx)+3sin(logx)x2=sin(logx)7cos(logx)x2Thus,d2ydx2=y2=sin(logx)7cos(logx)x2L.H.S.=x2y2+xy1+y=x2(sin(logx)7cos(logx)x2)+x(3sin(logx)+4cos(logx)x)+3cos(logx)+4sin(logx)=sin(logx)7cos(logx)3sin(logx)+4cos(logx)+3cos(logx)+4sin(logx)=4sin(logx)+4sin(logx)7cos(logx)+7cos(logx)=0=R.H.S.Hence proved.

Q.113

Ify=Aemx+Benx, show that d2ydx2(m+n)dydx+mny=0

Ans.

Given that y=A emx + BenxDifferentiating w.r.t. x, we get  dydx=Addxemx + Bddxenx=Amemx + BnenxAgaindifferentiating w.r.t. x, we getd2ydx2=Amddxemx + Bnddxenx=Am2emx + Bn2enxL.H.S.=d2ydx2(m+n)dydx+mny=(Am2emx+Bn2enx)(m+n)(Amemx+Bnenx)+mn(A emx + Benx)=Am2emx+Bn2enxAm2emxBmnenxAmnemxBn2enx+Amnemx+Bmnenx=0=R.H.S.

Q.114

Ify=500e7x+600e7x, show that d2ydx2= 49y

Ans.

Given that y=500 e7x + 600e7xDifferentiating w.r.t. x, we get      dydx=500ddxe7x +600ddxe7x    =500×7e7x+600×(7)e7x    =7(500e7x600 e7x)Againdifferentiating w.r.t. x, we get    d2ydx2=7(500×7e7x600×(7)e7x)d2ydx2=49(500e7x+600 e7x)d2ydx2=49yHence proved.

Q.15

If ey(x+1)=1, show that d2ydx2=(dydx)2

Ans.

Given that e y ( x+1 )=1 e y = 1 x+1 y=log( 1 x+1 ) y=log( x+1 ) Differentiating w.r.t. x, we get dy dx = d dx ( log( x+1 ) ) = d dx log( x+1 ) = 1 x+1 d dx ( x+1 ) = 1 x+1 ×( 1+0 ) = 1 x+1 Againdifferentiating w.r.t. x, we get d 2 y d x 2 = d dx ( 1 x+1 ) d 2 y d x 2 = d dx ( x+1 ) 1 d 2 y d x 2 =( 1 ) ( x+1 ) 2 d dx ( x+1 ) = 1 ( x+1 ) 2 = ( 1 x+1 ) 2 d 2 y d x 2 = ( dy dx ) 2 Hence proved. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@A490@

Q.116

If y=(tan1x)2, show that (x2+1)y2+2x(x2+1)y1=2

Ans.

The given function is y=(tan1x)2Differentiating both sides w.r.t. x, we get    dydx=ddx(tan1x)2=2tan1xddxtan1x              y1=2tan1x(11+x2)(1+x2)y1=2tan1xDifferentiatingagain w.r.t. x, we get        ddx{(1+x2)y1}=2ddxtan1x(1+x2)ddxy1+y1ddx(1+x2)=2×11+x2          (1+x2)y2+y1(0+2x)=2×11+x2  (1+x2)y2+2xy1=21+x2    (1+x2)2y2+2x(1+x2)y1=2(x2+1)2y2+2x(x2+1)y1=2Hence proved.

Q.117

VerifyRollesTheoremforthefunctionf(x)=x2+2x8,x[4,2].

Ans.

The given function, f(x)=x2+2x8, is a polynomial. So, it is continuous in [4,2] and is differentiable in (4,2).f(4)=(4)2+2(4)8  =1688=0  f(2)=(2)2+2(2)8  =448=0f(4)=f(2)=0The value of f(x) at 4 and 2 coincide.Rolles Theorem states that there is a point c(4, 2) such thatf(c)=0f(x)=x2+2x8Differentiating f(x) w.r.t. x, we getf(x)=ddx(x2+2x8)        =2x+2f(c)=2c+2∵f(c)=02c+2=0c=1(4, 2)Hence, Rolles Theorem is verified for the given function.

Q.118

Examine if Rolles Theorem is applicable to any of the following functions. Can you saysome thing about the converse of Rolles theorem from these example?ifx=xfor x5,9iifx=x    for x2,2iiifx=x21 for x1,2

Ans.

(i) Given function is f(x)=[x] for x[5, 9]Sincethe given function f(x) is not continuous at every integral point and in particular it is not continuous at x=5 and x=9.f (x) is not continuous in [5, 9].f(5)=[5]=5f(9)=[9]=9f(5)f(9)The differentiability of f in (5, 9) is checked as follows.Let n be an integer such that n(5, 9).The left hand limit of f at x=n islimh0f(n+h)f(n)h=limh0[n+h][n]h        =limh0n1nh[Since, h is very small number.So,limh0[n+h]=limh0(n1)Example:[50. 3]=[4. 7]=4]        =limh01h=The right hand limit of f at x=n islimh0+f(n+h)f(n)h=limh0+[n+h][n]h        =limh0+nnh[Since, h is very small number.So,limh0+[n+h]=limh0+(n)Example:[5+0.3]=[5.3]=5]        =limh0+0        =0Left Hand DerivativeRight Hand DerivativeSo, 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 Rolls Theorem.Hence, Rolls Theorem is not applicable for f(x)=[x] for x[5, 9](ii)Given function is f(x)=[x] for x[2, 2]Sincethe given function f(x) is not continuous at every integral point and in particular it is not continuous at x=2 and x=2.f (x) is not continuous in [2, 2].f(2)=[2]=2  f(2)=[2]=2f(2)f(2)The differentiability of f in (2, 2) is checked as follows.Let n be an integer such that n(2, 2).The left hand limit of f at x=n islimh0f(n+h)f(n)h=limh0[n+h][n]h        =limh0n1nh[Since, h is very small number.So,limh0[n+h]=limh0(n1)Example:[50.3] =[4.7]=4]        =limh01h= The right hand limit of f at x=n is lim h 0 + f( n+h )f( n ) h = lim h 0 + [ n+h ][ n ] h = lim h 0 + nn h [ Since, h is very small number. So, lim h 0 + [ n+h ]= lim h0+ ( n ) Example:[ 5+0.3 ]=[ 5.3 ]=5 ] = lim h 0 + 0 =0 Left Hand DerivativeRight 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[ 2,2 ] ( iii )The 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 =41 =3 f( 1 )f( 2 ) The 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 1forx[ 1,2 ] 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Q.119

If f: 5, 5Risdifferentiable function and if fx does not vanish anywhere, then prove that f 5f 5  

Ans.

Given function f:[5,5]R‹ is a differentiable function.Since every differentiable function is a continuous function thenwe have(a) f is continuous on [5, 5].(b) f is differentiable on (5, 5).Therefore, by the Mean Value Theorem, there exists c(5, 5) such that      f(c)=f(5)f(5)5(5)      =f(5)f(5)1010f(c)=f(5)f(5)It is also given that f(x)  does not vanish anywhere.            f(c)0      10f(c)0f(5)f(5)0            f(5)f(5)Hence, proved.

Q.120

VerifyMeanValueTheorem,iffx=x24x3inthe intervala, b,wherea= 1andb= 4.

Ans.

VerifyMeanValueTheorem,iffx=x24x3inthe intervala, b,wherea= 1andb= 4.

Q.121

Verify Mean Value Theorem,if f(x)=x35x2 – 3x in the interval [a,b],wherea= 1and b= 3. Find all c(1, 3)for which f‘(c) = 0.

Ans.

The given function=x35x23xSince, function f(x) is polynomial, so it is continuous in [1, 3]and it is differentiable in (1,3).Here,f(x)=ddx(x35x23x)  =3x210x3and,    f(c)=3c210c3Now,  f(1)=(1)35(1)23(1)  =153  =7and    f(3)=(3)35(3)23(3)  =27459  =27f(b)f(a)ba=f(3)f(1)31f(b)f(a)ba=27(7)2f(b)f(a)ba=202f(b)f(a)ba=10Mean Value Theorem states that there is a point c(1,3)such that f(c)=10∵        f(c)=103c210c+7=03c23c7c+7=03c(c1)c7(c1)=0(c1)(3c7)=0          c=1,  73, where c=73(1,3)Hence, Mean Value Theorem is verified for the given functionc=73(1,3)is the only point for whichf(c)=0.

Q.122

Examine the applicability of Mean Value Theorem for all threefunctions given below.(i)f(x)=[x]forx[5,9](ii)f(x)=[x]forx[2,2](iii)f(x)=x21forx[1,2]

Ans.

(i) Given function is f(x)=[x] for x[5,9]Sincethe given function f(x) is not continuous at every integral point and in particular it is not continuous at x=5 and x=9.f (x) is not continuous in [5, 9].f(5)=[5]=5f(9)=[9]=9f(5)f(9)The differentiability of f in (5, 9) is checked as follows.Let n be an integer such that n(5, 9).The left hand limit of f at x=n islimh0f(n+h)f(n)h=limh0[n+h][n]h        =limh0n1nh[Since, h is very small number.So,limh0[n+h]=limh0(n1)Example:[50.3]=[4.7]=4]        =limh01h=The right hand limit of f at x=n islimh0+f(n+h)f(n)h=limh0+[n+h][n]h        =limh0+nnh[Since, h is very small number.So,limh0+[n+h]=limh0+(n)Example:[5+0.3]=[5.3]=5]        =limh0+0        =0Left Hand DerivativeRight Hand DerivativeSo, 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[5,9](ii)Given function is f(x)=[x] for x[2,2]Sincethe given function f(x) is not continuous at every integral point and in particular it is not continuous at x=2 and x=2.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(2, 2).The left hand limit of f at x=n islimh0f(n+h)f(n)h=limh0[n+h][n]h        =limh0n1nh[Since, h is very small number.So,limh0[n+h]=limh0(n1)Example:[50.3]=[4.7]=4]        =limh01h=The right hand limit of f at x=n islimh0+f(n+h)f(n)h=limh0+[n+h][n]h        =limh0+nnh[Since, h is very small number.So,limh0+[n+h]=limh0+(n)Example:[5+0.3]=[5.3]=5]        =limh0+0        =0Left Hand DerivativeRight Hand DerivativeSo, 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[2,2] ( iii )The 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 1forx[ 1,2 ] It can be proved as: f( 1 )= ( 1 ) 2 1=0 f( 2 )= ( 2 ) 2 1=3 f( b )f( a ) ba = f( 2 )f( 1 ) 21 = 30 1 =3 ∵ f( x )=2x f( c )=3 2c=3 c= 3 2 =1.5[ 1,2 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@D192@

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

Ans.

Let y=(3x29x+5)9Differentiating w.r.t. x, we get  dydx=ddx(3x29x+5)9=9(3x29x+5)8ddx(3x29x+5)[By chain rule]=9(3x29x+5)8(6x9)=9×3(2x3)(3x29x+5)8=27(2x3)(3x29x+5)8

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.

Let  y=(5x)3cos2xTaking log both sides, we get      logy=log(5x)3cos2x      logy=3cos2x.log5xDifferentiating w.r.t. x, we getdydxlogy=3cos2xddxlog5x+3log5xddxcos2x      1ydydx=3cos2x(15xddx5x)+3log5x(sin2x)ddx2x[By chain rule]        dydx=y{35xcos2x(5)3log5xsin2x(2)}      =(5x)3cos2x{3cos2xx6sin2x log5x}

Q.126

Differentiate w.r.t. x the function:sin1(xx),0x1

Ans.

Let y=sin1(xx)Differentiating w.r.t. x, we get  dydx=ddxsin1(xx)=11(xx)2ddx(xx)[By chain rule]=11x3ddxx32=11x3×32x12=3x21x3=32x1x3

Q.127

Differentiate w.r.t. x the function:cos1(x2)2x+7,2<x<2

Ans.

Lety=cos1x22x+7,2<x<2Differentiating w.r.t.x, we get  dydx=ddxcos1x22x+7=2x+7ddxcos1x2cos1x2ddx2x+72x+72        By Quotientrule=2x+711x22ddxx2cos1x2122x+7ddx2x+72x+7  By chainrule=22x+74x2×12cos1x2122x+722x+7  dydx=2x+74x2cos1x212x+72x+7  =14x2×12x+7cos1x22x+732dydx=14x2×12x+7+cos1x22x+732

Q.128

Differentiate w.r.t. x the function: co t 1 [ 1+sinx + 1sinx 1+sinx 1sinx ],0x π 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaaeaqbaaGceaqabeaacaWHebGaaCyAaiaahAgacaWHMbGaaCyzaiaahkhacaWHLbGaaCOBaiaahshacaWHPbGaaCyyaiaahshacaWHLbGaaeiiaiaahEhacaGGUaGaaCOCaiaac6cacaWH0bGaaiOlaiaabccacaWH4bGaaeiiaiaahshacaWHObGaaCyzaiaabccacaWHMbGaaCyDaiaah6gacaWHJbGaaCiDaiaahMgacaWHVbGaaCOBaiaacQdaaeaaieqacaWFJbGaa83Baiaa=rhadaahaaWcbeqaaiabgkHiTiaaigdaaaGcdaWadaqaamaalaaabaWaaOaaaeaacaWFXaGaa83kaiaa=nhacaWFPbGaa8NBaiaa=HhaaSqabaGccaWFRaWaaOaaaeaacaWFXaGaeyOeI0Iaa83Caiaa=LgacaWFUbGaa8hEaaWcbeaaaOqaamaakaaabaGaa8xmaiaa=TcacaWFZbGaa8xAaiaa=5gacaWF4baaleqaaOGaeyOeI0YaaOaaaeaacaWFXaGaeyOeI0Iaa83Caiaa=LgacaWFUbGaa8hEaaWcbeaaaaaakiaawUfacaGLDbaacaGGSaGaaGPaVlaa=bdacqGHKjYOcaWF4bGaeyizIm6aaSaaaeaarmWu51MyVXgaiyaacqGFapaCaeaacaaIYaaaaaaaaa@87AD@

Ans.

Let cot1[1+sinx+1sinx1+sinx1sinx],  0xπ2Now,  1+sinx+1sinx1+sinx1sinx=1+sinx+1sinx1+sinx1sinx×1+sinx+1sinx1+sinx+1sinx=(1+sinx+1sinx)2(1+sinx)2(1sinx)2=1+sinx+1sinx+21+sinx1sinx1+sinx(1sinx)=2+21sin2x1+sinx1+sinx=2+2cosx2sinx=2(1+cosx)2sinx=2cos2x22sinx2cosx2=cotx2Differentiating w.r.t. x, we get  dydx=ddxcot1[1+sinx+1sinx1+sinx1sinx]=ddxcot1[cotx2][By Quotient rule]=ddx(x2)=12.Thus, dydx=12.

‘Q.129

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

Ans.

Let     y=logxlogx, x>1Taking log both sides, we get      logy=loglogxlogx      logy=logx.loglogxDifferentiating w.r.t. x, we getdydxlogy=logxddxloglogx+loglogxddxlogx  By product rule  1ydydx=logx1logxddxlogx+loglogx1xBy chain rule      dydx=y1x+1xloglogx     =logxlogx1x+1xloglogx

Q.130

Differentiate w.r.t.x the function:cos(acosx+bsinx), for some constant a and b.

Ans.

Let y=cos (acosx+bsinx)Differentiating w.r.t. x, we get  dydx=ddxcos (acosx+bsinx)=sin(acosx+bsinx)ddx(acosx+bsinx)[By chain rule]=sin(acosx+bsinx)×(asinx+bcosx)=(bcosxasinx)sin(acosx+bsinx)

Q.131

Differentiate w.r.t.x the function:(sinxcosx) (sinxcosx),π4< x <3π4

Ans.

Let  y=(sinxcosx)(sinxcosx), π4<x<3π4Taking log both sides, we get      logy=log(sinxcosx)(sinxcosx)      logy=(sinxcosx).log(sinxcosx)Differentiating w.r.t. x, we getdydxlogy=(sinxcosx)ddxlog(sinxcosx)+log(sinxcosx)ddx(sinxcosx)  [By product rule]      1ydydx=(sinxcosx)(1(sinxcosx)ddx(sinxcosx))        +log(sinxcosx).(cosx+sinx)      [By chain rule]        dydx=y[(sinxcosx){1(sinxcosx)(cosx+sinx)}+(cosx+sinx)log(sinxcosx)]        dydx=(sinxcosx)(sinxcosx)(cosx+sinx){1+log(sinxcosx)}

Q.132

Differentiate w.r.t.x the function:xx+xa+ax+aa,for some fixed a>0 and x>0

Ans.

Let  y=xx+xa+ax+aaDifferentiating w.r.t. x, we get  dydx=ddx(xx+xa+ax+aa)=ddxxx+ddxxa+ddxax+ddxaa=ddxxx+axa1+axlogea+0=ddxxx+axa1+axlogea...(i)Letu=xxTaking log both sides, we getlogu=logxx=xlogxDifferentiating w.r.t. x, we getddxlogu=ddx(xlogx)    1ududx=xddxlogx+logxddxx        dudx=u{x.1x+logx.(1)}        dudx=xx(1+logx)From equation(i), we get  dydx=xx(1+logx)+axa1+axlogea

Q.133

Differentiate w.r.t. x the function:xx23+(x3)x2, for x>3

Ans.

Lety=xx23+(x3)x2Letu=xx23,  v=(x3)x2So,  y=u+vdydx=dudx+dvdx  ...(i)Taking log both sides, we getlogu=logxx23=(x23)logxandlogv=log(x3)x2=x2log(x3)Differentiating w.r.t. x, we getddxlogu=ddx{(x23)logx}    1ududx=(x23)ddxlogx+logxddx(x23)        dudx=u{(x23).1x+logx.(2x)}        dudx=x(x23){(x23)x+logx.(2x)}andddxlogv=ddx{x2log(x3)}    1vdvdx=x2ddxlog(x3)+log(x3)ddxx2        dvdx=v{x2.1x3+log(x3).(2x)}        dvdx=(x3)x2{x2x3+2x.log(x3)}From equation(i), we get  dydx=x(x23){(x23)x+2x.logx}+(x3)x2{x2x3+2x.log(x3)}

Q.134

Find dydx, if y=12 (1cost), x=10 (tsint), π2 < t < π2

Ans.

It is given that, y=12(1cost),x=10(tsint)Differentiating w.r.t. t, we getdydt=ddt12(1cost)  and  dxdt=ddt10(tsint)dydt=12(0+sint)  and      dxdt=10(1cost)dydt=12sint  and      dxdt=10(1cost)    dydx=dydtdxdt    =12sint10(1cost)dydx=6×2sint2cost25×2sin2t2dydx=65cott2.

Q.135

Finddydx,ify=sin1x+sin11x2,1x1

Ans.

Let y=sin1x+ sin11x2,1x1Differentiating w.r.t. x, we get  dydx=ddx(sin1x+sin11x2)=ddxsin1x +ddxsin11x2=11x2+11(1x2)2ddx(1x2)  [By chain rule]=11x2+11(1x2)×121x2ddx(1x2)=11x2+1x2×121x2(02x)=11x2+1x×121x2(2x)=11x211x2dydx=0.

Q.136

If x 1+y + y1+x =0, for, 1<x<1, prove thatdydx=1(1+x)2

Ans.

Given equation is x1+y +y1+x=0      x1+y=y1+xSquarring both sides, we get        x2(1+y)=y2(1+x)x2+x2y=y2+xy2  x2y2=xy2x2y    (xy)(x+y)=xy(yx)    (xy)(x+y)=xy(xy)(x+y)=xy  y+xy=x        (1+x)y=x  y=x1+xDifferentiating both sidesw.r.t. x, we get          dydx=ddxx(1+x)        dydx={(1+x)ddxxxddx(1+x)(1+x)2}          dydx={(1+x).1x(0+1)(1+x)2}          dydx={1+xx(1+x)2}        dydx=1(1+x)2Henceproved.

Q.137

If (xa)2+(yb)2=c2, for some c>0, prove that[1+(dydx)2]32d2ydx2is a constant independent of a and b.

Ans.

Given equation is   (xa)2+(yb)2=c2Differentiating both sidesw.r.t. x, we get          ddx(xa)2+ddx(yb)2=ddxc22(xa)ddx(xa)+2(yb)ddx(yb)=0 [By chain rule]          2(xa)(10)+2(yb)(dydx0)=0          2(xa)+2(yb)dydx=0          dydx=xayb  ...(i)Differentiating both sides again w.r.t. x, we get          d2ydx2=ddx(xayb)[Byquotient  rule]        =(yb)ddx(xa)(xa)ddx(yb)(yb)2        =(yb)(10)(xa)(dydx0)(yb)2        =(yb)(xa)(xayb)(yb)2[Fromequation (i)]      d2ydx2=(yb)(xa)(xayb)(yb)2    =(yb)2+(xa)2(yb)3d2ydx2=c2(yb)3[∵(xa)2+(yb)2=c2]Now,  [1+(dydx)2]32d2ydx2=[1+(xayb)2]32c2(yb)3    ={(yb)2+(xa)2(yb)2}32c2(yb)3    ={c2(yb)2}32c2(yb)3    =c3(yb)3c2(yb)3    =c,which is a contant and is independent of a and b.Hence proved.

Q.138

If cos y = x cosa+y, with cos a ±âŸ1, prove that     dydx=cos2a+ysina

Ans.

Given equation is cosy=xcos(a+y) ...(i)Differentiating both sides with respect to x, we get    ddxcosy=ddx{xcos(a+y)}sinydydx=xddxcos(a+y)+cos(a+y)ddxx=x×sin(a+y)ddx(a+y)+cos(a+y)×1=xsin(a+y)dydx+cos(a+y){xsin(a+y)siny}dydx=cos(a+y)          {cosycos(a+y)sin(a+y)siny}dydx=cos(a+y)  [Putting value of x from equation(i)]{cosysin(a+y)sinycos(a+y) }dydx=cos2(a+y)        sin(a+yy)dydx=cos2(a+y)                       sinadydx=cos2(a+y)                              dydx=cos2(a+y)sinaHence  proved.

Q.139

If x= a(cost+t sint) andy=a(sinttcost), find d2ydx2.

Ans.

Given equations arex=a(cost+tsint) and y=a(sintt cost)Differentiating w.r.t. t, we get  dxdt=ddta(cost+tsint)=a(sint+tddtsint+sintddtt)=a(sint+tcost+sint)=atcostdydt=ddta(sintt cost)=a(costtddtcostcostddtt)=a(cost+tsintcost)=atsintdydx=dydtdxdt=atsintatcost=tantAgain differentiating w.r.t. x, we getd2ydx2=ddxtant=sec2tdtdx=sec2t×1atcost[∵dxdt=atcostdtdx=1atcost]=sec3tat,0<t<π2

Q.140

If f(x)=|x|3,show that f(x)exists for all realx,and findit.

Ans.

Since,  |x|={x, ifx0x, ifx<0Therefore, when x0, f(x)=|x|3=x3So, f(x)=3x2 and f(x)=6xwhen x<0, f(x)=|x|3=(x)3=x3So,f(x)=3x2 and f(x)=6xThus,for f(x)=|x|3,f(x) exists for all real x and is given by,f(x)={6x, ifx06x, ifx<0

Q.141

Using mathematical induction prove thatddx(xn)=nxn1 for all positive integersn.

Ans.

LetP(n):ddx(xn)=nxn1for all positive integers nFor n = 1,P(1):  ddx(x1)=1=1x11P(n) is true for n = 1Let P(k) is true for some positive integer k.P(k):ddx(xk)=kxk1  ...(i)It should be true for n=k+1So,P(k+1):ddx(xk+1)=(k+1)x(k+1)1P(k+1):ddx(xk+1)=(k+1)xkL.H.S.=ddx(xk+1)=ddx(xk.x)=xkddxx+xddxxk=xk+x.kxk1=xk+kxk=(1+k)xk=(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.

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.

Given equation is          sin(A+B)=sinAcosB+cosAsinBDifferentiating both sides with respect to x, we get            ddxsin(A+B)=ddx(sinAcosB)+ddx(cosAsinB)cos(A+B)ddx(A+B)=sinAddxcosB+(ddxsinA)cosB+cosAddxsinB+(ddxcosA)sinBcos(A+B)(dAdx+dBdx)=sinA(sinBdBdx)+(cosAdAdx)cosB+cosAcosBdBdx+(sinAdAdx)sinBcos(A+B)(dAdx+dBdx)=sinAsinBdBdx+cosAdAdxcosB+cosAcosBdBdxsinAdAdxsinBcos(A+B)(dAdx+dBdx)=(cosAcosBsinAsinB)dAdx+(cosAcosBsinAsinB)dBdxcos(A+B)(dAdx+dBdx)=(cosAcosBsinAsinB)(dAdx+dBdx)      cos(A+B)=(cosAcosBsinAsinB)

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

Ans.

Yes, there exists a function which is continuous everywherebut not differentiable at exactly two points.For example:f(x)=|x1|+|x2|,xR  is continuous at everywhere but not differentiable at x=1 and 2.f(x)=|x1|+|x2|={(x1)(x2),  ifx<1(x1)(x2),if  1x<2(x1)+(x2),if  x2={2x+3,ifx<11,if  1x<22x3,if  x2Since, f is continuous at all points.So, differentiability of f at x=1L.H.D.=limh0f(1+h)f(1)h    =limh02(1+h)+3{2(1)+3}h    =limh022h+3+23h        =limh02hh      =2R.H.D.=limh0+f(1+h)f(1)h    =limh0+11h    =limh0+0h    =0L.H.D.R.H.D.Thus,function is not differentiable at x=1.Now, differentiability of f at x=2L.H.D.=limh0f(2+h)f(2)h    =limh011h    =limh00h    =0R.H.D.=limh0+f(2+h)f(2)h    =limh0+2(2+h)31h    =limh0+4+2h4h    =2L.H.D.R.H.D.Thus, function f is not differentiable at x=2.Therefore, function f is continuous at everywhere but it is notdifferentiable atexactly two points x=1 and 2.

Q.144

If y = | f ( x ) g ( x ) h ( x ) l m n a b c | , p r o v e t h a t d y d x = | f ( x ) g ( x ) h ( x ) l m n a b c |

Ans.

W e h a v e y = | f ( x ) g ( x ) h ( x ) l m n a b c | = f ( x ) | m n b c | + g ( x ) | n l c a | + h ( x ) | l m a b | 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 d y d x = d d x f ( x ) | m n b c | + d d x g ( x ) | n l c a | + d d x h ( x ) | l m a b | d y d x = f ( x ) | m n b c | + g ( x ) | n l c a | + h ( x ) | l m a b | = | f ( x ) g ( x ) h ( x ) l m n a b c | T h e r e f o r e , d y d x = | f ( x ) g ( x ) h ( x ) l m n a b c |

Q.145

Ify=eacos1x,1x1,show that 1x2d2ydx2xdydxa2y=0.

Ans.

Given that, y=eacos1xTaking logarithm both sides, we get        logy=logeacos1x        =acos1xloge        =acos1x×1[∵loge=1]Differentiating both sides,with respect to x, we get    ddxlogy=ddxacos1x    1ydydx=a11x2        dydx=ay1x2Squarring both sides, we get      (dydx)2=(ay1x2)2  (dydx)2=a2y2(1x2)(1x2)(dydx)2=a2y2Differentiating again both sides,with respect to x, we get(1x2)ddx(dydx)2+(dydx)2ddx(1x2)=a2ddxy2(1x2).2(dydx)d2ydx22x(dydx)2=2a2ydydx2(dydx){(1x2)d2ydx2x(dydx)}=2a2ydydx(1x2)d2ydx2x(dydx)=a2y[∵dydx0](1x2)d2ydx2x(dydx)a2y=0Hence proved.

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FAQs (Frequently Asked Questions)

1. What are the vital topics covered in NCERT Solutions Class 12 Mathematics Chapter 5?

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. 

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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