# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.2) Exercise 5.2

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 play an important role in the preparation for the CBSE Board Examinations for students of Class 12. The Class 12 Boards play a determining role in building the career of students. The chapter on Continuity and Differentiability is one of the most important chapters for the examinations. Often, multiple questions are asked in the CBSE Board Examinations from this chapter. The CBSE recommends NCERT textbooks for its examinations. Questions from the NCERT exercises are frequently asked in exams. The NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.2 make the task of students much easier as they prepare for the board examinations. Chapter 5 of the Class 12 NCERT textbook deals with the concepts of Continuity and Differentiability. While Exercise 5.1 of the NCERT textbook is based on the concept of Continuity, Exercise 5.2 Class 12th is based on the concept of Differentiability. Questions are also based on the concept of Derivatives Of Composite Functions. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 by Extramarks provide easy-to-understand solutions to the questions asked in this exercise. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 have been prepared by experienced subject experts keeping in view the level of understanding of the students of Class 12. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 have been prepared according to the latest syllabus prescribed by the CBSE for Class 12. Therefore, students need not worry about the syllabus and look elsewhere. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 serve as the one-stop destination for students of Class 12 to find easy and authentic solutions to all the questions of Class 12 Maths Chapter 5 Exercise 5.2.

## NCERT Maths Class 12 Exercise 5.2 Solutions

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 cover the topics of differentiability and derivatives of composite functions. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 cover the chain rule which is essential throughout the chapter and beyond in finding derivatives of functions. Exercise 5.2 contains 10 questions. All these questions are based on the differentiability of a given function. Eight of these questions ask the students to differentiate a function with respect to a variable x. Some of these functions include trigonometric functions. Students should be able to remember the derivatives of trigonometric functions. These problems can be solved by using the chain rule discussed before the exercise. The remaining two questions ask the students to test the differentiability of the given functions at certain points. All these questions are short and long answer questions. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 provide solutions to all the questions in detail. Practising NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 multiple times will guide students to score maximum marks in their examinations as well as solve a lot of other problems related to differentiation.

### List Of Topics Covered Under NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

The topics covered under NCERT Solutions for Class 12 Mathematics Chapter 5 are given in the following table.

 Continuity Algebra of continuous functions Differentiability Derivatives of composite functions Derivatives of implicit functions Derivatives of inverse trigonometric functions Exponential and Logarithmic Functions Logarithmic Differentiation Derivatives of Functions in Parametric Forms Second Order Derivative Mean Value Theorem

### List of Important Conditions/Theorems in NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.2

• Differentiability:- The differentiability test of a function determines if a derivative of the function exists at every point in its domain. A function is said to be differentiable in an interval (a, b) if it is differentiable at every point in the interval (a, b), i.e., h0f(c+h)-f(c)h  exists for every point c in the interval (a, b).
• A function must be continuous in order to be differentiable.
• Chain rule:– The chain rule is used to find the derivative of a composite function. According to the chain rule, the derivative of f(g(x)) is f'(g(x))⋅g'(x)

### Chapter 5 Continuity and Differentiability Class 12 Maths

The concepts of Continuity and Differentiability are two of the most important concepts for students not only in Mathematics but also in other subjects like Physics. So, thoroughly practising the questions in Class 12 Maths Ch 5 Ex 5.2 helps students solve many other questions related to these concepts. The chapter starts with the concept of continuity, where students learn to check the continuity of a function. Next, students learn some algebraic continuous functions. Then, they learn to check the differentiability of a function. Students also learn to find the derivative of a composite function using the chain rule. Further, they are also taught to find the derivative of implicit functions and inverse trigonometric functions. Students also learn the concepts of exponential and logarithmic functions and differentiate a function using a logarithm. They also learn to find derivatives of functions in parametric form. After that, the second-order derivative is introduced. The chapter ends with Rolle’s Theorem and Mean Value Theorem. The concepts introduced in this chapter can be applied in different fields. Differentiation is applied in finding the rate of change of one entity with respect to another entity like speed, acceleration, etc. This chapter is also very essential for students preparing for engineering entrance examinations like JEE Mains and JEE Advanced. Practising the NCERT questions repetitively helps students easily solve the problems in the examinations. Students can also refer to the study material provided by Extramarks along with the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 for an easier understanding of the chapter. Differentiation is especially important for students who wish to pursue higher studies in Mathematics, Physics and Engineering. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 can be of great help for them as they will be acquainted with finding the derivatives of various types of functions that come their way.

### NCERT Math Solution Class 12 Chapter 5 Exercise 5.2

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 have been prepared by subject matter experts in such a way that students find it easy and enjoyable to read. Exercise 5.2 is one of the easiest exercises in the chapter. Students can easily answer the questions in this exercise when they solve these questions according to the step-by-step solutions provided in NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2. Once students thoroughly solve this exercise on their own, they will be able to solve all the questions based on checking the differentiability of a function using the addition, subtraction, product, and quotient rules. They can also differentiate the functions using the chain rule. Sometimes students are unable to identify composite functions and hence get confused while solving problems. By referring to the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2, students get acquainted with the various types of functions in which chain rule can be applied to differentiate. Ultimately, they are able to differentiate a lot of functions and answer the questions confidently.

### NCERT Maths Class 12 Exercise 5.2 Solutions – Weightage of Marks

The chapter on Continuity and Differentiability is a part of the Calculus unit. As per the latest syllabus released by the CBSE, this unit has been assigned 35 out of 80 marks in the board examinations. Hence, it has more weightage than any other unit. Students need to be thorough in this chapter to score higher marks. The syllabus prescribes 20 periods for this chapter. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 are prepared considering the topic weightage as per the latest CBSE syllabus. Most of the questions in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 are short answer-type questions that carry 2-3 marks in the examinations. Sometimes multiple choice questions are also asked in this section. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 by Extramarks have been prepared accordingly.

### Benefits of NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.2

• NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 are beneficial to students preparing for board examinations as they are solved in a simple and easy-to-understand manner that every student can refer to.
• The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 are written stepwise that help in securing complete marks in the examinations.
• NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 are prepared according to the latest syllabus and guidelines prescribed by the CBSE so that students do not have to look elsewhere for the concepts included in the syllabus or the marks distribution.
• NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 by Extramarks provide all the required answers to the questions asked in the exercise and are completely error-free.
• Practising the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 multiple times will help the students attempt such questions quickly and using the correct stepwise approach in the examinations.

Students can use the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 anytime and anywhere without the need for any external guidance as they are written in a simple way.

### NCERT Solutions for Class 12 Maths Chapter 5 Other Exercises

 Chapter 5 – Continuity and Differentiability Exercises Exercise 5.1 10 Short Questions and 24 Long Questions Exercise 5.3 9 short Questions and 6 long Questions Exercise 5.4 5 Short Questions and 5 Long Questions Exercise 5.5 4 Short Questions and 14 Long Questions Exercise 5.6 1 Short Question and 1 Long Question Exercise 5.7 10 Short Questions and 7 Long Questions Exercise 5.8 Questions with Solutions

Q.1 Differentiate the function with respect to x.

sin(x2 + 5)

Ans

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

Q.2 Differentiate the function with respect to x.

cos(sinx)

Ans

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

Q.3 Differentiate the function with respect to x.

sin (ax+b)

Ans

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

Q.4 Differentiate the function with respect to x.

$\text{sec}\text{(tan}\text{(}\sqrt{\mathrm{x}}\right)\right)$

Ans

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

Q.5

$\text{Differentiate the functions with respect to x}\frac{sin\left(ax+\mathrm{b}\right)}{cos\left(cx+\mathrm{d}\right)}$

Ans

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

Q.6 Differentiate the function with respect to x.

cosx3. sin2(x5)

Ans

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

Q.7

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

Ans

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

Q.8

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

Ans

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

Q.9

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

Ans

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

Q.10

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

Ans

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

## 1. How many questions are included in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2?

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 contain solutions to 10 questions from Exercise 5.2.

## 2. What are the topics covered in NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2?

NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 cover the topics of Differentiability of a Function and Derivatives of a Cmposite Function.

## 3. Do NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 help in the preparation for the CBSE Board Examinations?

NCERT Solutions are the most useful part in the preparation for the CBSE Board Examinations. The NCERT Mathematics textbook is recommended by the CBSE for its board examinations, and most questions are based on the NCERT textbook. The step-by-step solutions in NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2 help in securing full marks for these questions.

## 4. What kind of questions does Exercise 5.2 include?

Exercise 5.2 includes questions that ask the students to differentiate the given functions concerning variable x. Certain questions are there to prove that a given function is not differentiable at a certain point in its domain.

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