# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.4) Exercise 5.4

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.4) Exercise 5.4

Vedantu gives a whole new world to the way you learn and study in 12th Class. With the help of our ex-5.4 Class 12 NCERT Solutions, you can learn faster, and in an effective way to get things done in the right direction. You can access our Free PDF of NCERT Solutions for ex 5.4 Class 12 prepared by our professional teachers as per CBSE guidelines. This way, you can revise the entire exercise in a limited period of time and score higher marks in your upcoming examinations.

Do you need help with your Homework? Are you preparing for Exams?
Study without Internet (Offline)

## Access NCERT Solutions for Class 12 Maths Chapter-5 Continuity and Differentiability

Exercise 5.4

1. Differentiating the following w.r.t. $x:\frac{{{e^x}}}{{\sin x}}$

Ans:

Let $y = \frac{{{e^x}}}{{\sin x}}$

Now differentiating w.r.t. $x$ , we get

$\frac{{dy}}{{dx}} = \frac{{\sin x \times \tfrac{d}{{dx}}({e^x}) - {e^x} \times \tfrac{d}{{dx}}(\sin x)}}{{{{\sin }^2}x}}$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{\sin x \times {e^x} - {e^x} \times \cos x}}{{{{\sin }^2}x}}$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^x}(\sin x - \cos x)}}{{{{\sin }^2}x}},x \ne n\pi ,n \in \mathbb{Z}$

2. Differentiating the following ${e^{{{\sin }^{ - 1}}x}}$

Ans:

Let $y = {e^{{{\sin }^{ - 1}}x}}$

Now differentiating w.r.t. $x$ , we get

$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^{{{\sin }^{ - 1}}x}}} \right)$

$\Rightarrow \frac{{dy}}{{dx}} = {e^{{{\sin }^{ - 1}}x}} \times \frac{d}{{dx}}({\sin ^{ - 1}}x)$

$\Rightarrow \frac{{dy}}{{dx}} = {e^{{{\sin }^{ - 1}}x}} \times \frac{1}{{\sqrt {1 - {x^2}} }}$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }},x \in ( - 1,1)$

3. Differentiating the following w.r.t. $x:{e^{{x^3}}}$

Ans: Let $y = {e^{{x^3}}}$

By using the quotient rule, we get

$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right)$

$\Rightarrow \frac{{dy}}{{dx}} = {e^{{x^3}}} \times \frac{d}{{dx}}\left( {{x^3}} \right)$

$\Rightarrow \frac{{dy}}{{dx}} = {e^{{x^3}}} \times 3{x^2}$

$\Rightarrow \frac{{dy}}{{dx}} = 3{x^2}{e^{{x^3}}}$

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

Ans:

Let $y = \sin ({\tan ^{ - 1}}{e^{ - x}})$

By using the chain rule we get

$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sin ({{\tan }^{ - 1}}{e^{ - x}})} \right]$

$\Rightarrow \frac{{dy}}{{dx}} = \cos ({\tan ^{ - 1}}{e^{ - x}}) \times \frac{d}{{dx}}({\tan ^{ - 1}}{e^{ - x}})$

$\Rightarrow \frac{{dy}}{{dx}} = \cos ({\tan ^{ - 1}}{e^{ - x}}) \times \frac{1}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}}\frac{d}{{dx}}({e^{ - x}})$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{\cos ({{\tan }^{ - 1}}{e^{ - x}})}}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}} \times {e^{ - x}} \times \frac{d}{{dx}}( - x)$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{ - x}}\cos ({{\tan }^{ - 1}}{e^{ - x}})}}{{1 + {e^{ - 2x}}}} \times ( - 1)$

$\Rightarrow \frac{{dy}}{{dx}} = - \frac{{{e^{ - x}}\cos ({{\tan }^{ - 1}}{e^{ - x}})}}{{1 + {e^{ - 2x}}}}$

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

Ans:

Let $y = \log (\cos {e^x})$

By using the chain rule, we get

$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\log (\cos {e^x})} \right]$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\cos {e^x}}} \times \frac{d}{{dx}}\left( {\cos {e^x}} \right)$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\cos {e^x}}} \times ( - \sin {e^x}) \times \frac{d}{{dx}}\left( {{e^x}} \right)$

$\Rightarrow \frac{{dy}}{{dx}} = - \frac{{\sin {e^x}}}{{\cos {e^x}}} \times \left( {{e^x}} \right)$

$\Rightarrow \frac{{dy}}{{dx}} = - {e^x}\tan {e^x},{e^x} \ne (2n + 1)\frac{\pi }{2},n \in \mathbb{N}$

6. Differentiating the following w.r.t. $x:{e^x} + {e^{{x^2}}} + ... + {e^{{x^5}}}$

Ans:

Let $y = {e^x} + {e^{{x^2}}} + ... + {e^{{x^5}}}$

$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^x} + {e^{{x^2}}} + ... + {e^{{x^5}}}} \right)$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^x}} \right) + \frac{d}{{dx}}\left( {{e^{{x^2}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^4}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^5}}}} \right)$

$\Rightarrow \frac{{dy}}{{dx}} = {e^x} + \left[ {{e^{{x^2}}}\frac{d}{{dx}}\left( {{x^2}} \right)} \right] + \left[ {{e^{{x^3}}}\frac{d}{{dx}}\left( {{x^3}} \right)} \right] + \left[ {{e^{{x^4}}}\frac{d}{{dx}}\left( {{x^4}} \right)} \right] + \left[ {{e^{{x^5}}}\frac{d}{{dx}}\left( {{x^5}} \right)} \right]$

$\Rightarrow \frac{{dy}}{{dx}} = {e^x} + \left[ {{e^{{x^2}}} \times 2x} \right] + \left[ {{e^{{x^3}}} \times 3{x^2}} \right] + \left[ {{e^{{x^4}}} \times 4{x^3}} \right] + \left[ {{e^{{x^5}}} \times 5{x^4}} \right]$

$\Rightarrow \frac{{dy}}{{dx}} = {e^x} + 2x{e^{{x^2}}} + 3{x^2}{e^{{x^3}}} + 4{x^3}{e^{{x^4}}} + 5{x^4}{e^{{x^5}}}$

7. Differentiating the following w.r.t. $x:\sqrt {{e^{\sqrt x }}} ,x > 0$

Ans:

Let $y = \sqrt {{e^{\sqrt x }}}$

Then ${y^2} = {e^{\sqrt x }}$

By differentiating the above equation with respect to $x$ , we get

${y^2} = {e^{\sqrt x }}$

$\Rightarrow 2y\frac{{dy}}{{dx}} = {e^{\sqrt x }}\frac{d}{{dx}}\left( {\sqrt x } \right)$      (By applying the chain rule)

$\Rightarrow 2y\frac{{dy}}{{dx}} = {e^{\sqrt x }} \times \frac{1}{2} \times \frac{1}{{\sqrt x }}$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4y\sqrt x }}$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4\sqrt {{e^{\sqrt x }}} \sqrt x }}$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4\sqrt {x{e^{\sqrt x }}} }},x > 0$

8. Differentiating the following w.r.t. $x:\log (\log x),x > 1$

Ans:

Let $y = \log (\log x)$

By using the chain rule in the above equation, we get

$\Rightarrow \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\log (\log x)} \right]$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\log x}} \times \frac{d}{{dx}}(\log x)$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\log x}} \times \frac{1}{x}$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{x\log x}},x > 1$

9. Differentiating the following w.r.t. $x:\frac{{\cos x}}{{\log x}},x > 0$

Ans:

Let $y = \frac{{\cos x}}{{\log x}}$

By using the quotient rule, we obtain

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{\tfrac{d}{{dx}}(\cos x) \times \log x - \cos x \times \tfrac{d}{{dx}}(\log x)}}{{{{\left( {\log x} \right)}^2}}}$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{( - \sin x)\log x - \cos x \times \frac{1}{x}}}{{{{(\log x)}^2}}}$

$\Rightarrow \frac{{dy}}{{dx}} = - \frac{{x\log x \times \sin + \cos x}}{{x{{(\log x)}^2}}},x > 0$

10. Differentiating the following w.r.t. $x:\cos (\log x + {e^x}),x > 0$

Ans:

Let $y = \cos (\log x + {e^x})$

By using the chain rule in the above equation, we get

$y = \cos (\log x + {e^x})$

$\frac{{dy}}{{dx}} = - \sin (\log x + {e^x}) \times \frac{d}{{dx}}(\log x + {e^x})$

$= - \sin (\log x + {e^x}) \times \left[ {\frac{d}{{dx}}(\log x) + \frac{d}{{dx}}({e^x})} \right]$

$= - \sin (\log x + {e^x}) \times \left( {\frac{1}{x} + {e^x}} \right)$

$= - \left( {\frac{1}{x} + {e^x}} \right)\sin (\log x + {e^x}),x > 0$

## Introduction To Vedantu’s Ex 5.4 Class 12 Maths NCERT Solutions

Presently, we are in the era of digitalization, where there are so many opportunities to make the studying process more lucrative. Technology and education are two faces of the same coin which can’t be separated from one another. To carve a great learning process, Vedantu has designed online PDF notes for ex-5.4 class 12 NCERT solutions as per the latest NCERT rules.

From solutions for every question in exercise to examples, and step by step description, you will learn every single thing that will make this Continuity and Differentiability concept more clear and understandable for you. Whether you just want to revise the 5.4 maths class 12 exercise or wish to study the complete concept completely again, Vedantu’s tailored NCERT Maths Solutions will surely help you attain higher marks in your exams.

### NCERT Ex 5.4 Class 12 Maths Solutions For You & Your Needs

Since the internet has completely revolutionized the ways of studying and is now more specific for everyone’s needs; there are so many benefits you can avail. Many students lookout for Exercise 5.4 Class 12 Maths solutions either for the purpose of revising or studying the concepts gain, on the internet.

To get these NCERT Exercise 5.4 class 12 solutions with Vedantu is the right choice you may make to get all the solutions to your answers. These answers are available in the PDF format, which can easily be availed from the internet without costing you any money.

### NCERT Solutions For Class 12 Maths Ex 5.4 In PDF Format

Continuity and Differentiability Exercise 5.4 Class 12 Maths solutions are prepared and designed in the PDF format by the best subject teachers of Vedantu. All the essential concepts and theories in Exercise 5.4 come with a detailed explanation which will make you understand the concepts in a better way. With these CBSE Class 12 Maths solutions, it will become more straightforward and more effortless for you to prepare well for the exams.

### Access Vedantu's Answers of Maths NCERT Class 12 Exercise 5.4 Continuity and Differentiability

Vedantu’s notes come with different modules to cover all the important and essential concepts that you would need to learn for the preparation of Exercise 5.4 maths class 12 for your final examinations. Our PDF solutions will comprise of:

• Answers to every question present in the exercise

• Step to step description for every step

• Revise notes from our experienced faculty

• Tips and suggestions for the final preparations

Obviously, in order to perform best in Class 12 maths exams, you need to understand the complete Continuity and Differentiability concept. Since class 12 is going to decide many things later on. Class 12th is considered to be one of the most crucial years of school life.

Marks of this class are later on going to decide which educational institute you would get enrolled in. Therefore, it gets vital for you to learn all the below-mentioned topics to ensure you are cleared with the complete concept of chapter 5 and exercise 5.4. So, make sure you know:

• The entire concepts of used technique

• Logic to solve the questions of exercise 5.4

• Understanding of Continuity and Differentiability concept

• Important points present in the chapter

• Methods to solve each exercise question with right steps

With the Vedantu’s NCERT Solution notes for Maths Class 12 Exercise 5.4, you can learn all these important as well as the above mentioned things in just a matter of few notes. Our notes are going to work great for you.

Just make sure to go through them once and then you will definitely be able to clear out all your concepts. These PDF notes and Exercise 5.4 solutions are going to work great for you.

### Why To Follow Up With Vedantu’s Ex-5.4 Class 12 NCERT Solutions

There are going to be plenty and tremendous reasons for you to opt for Vedantu’s Class 12 Continuity and Differentiability solutions. At one or some another part these reasons are definitely going to work for you. These notes could be the “key” for you to give your very best in the board examinations of class 12th.

Whether you are a student who is having a bit of knowledge about chapter 5 exercise 5.4 exercise or someone who have no clue of this exercise or someone who just wants brief notes to revise this exercise, getting the right notes and guidance will certainly help you to excel in the class 12 board examinations.

Vedantu’s highly professional and expert teachers, having diverse knowledge and skill sets in Maths subject, have brought a great opportunity for you in every possible case. With this NCERT exercise notes, you can perform your best in the final Maths examination paper and score higher marks. Apart from the other crowd, Vedantu has gained tremendous popularity for its quality of notes, which can really be a great help for you.

There are many attractive features of Vedantu’s NCERT Solutions for (Ex 5.4) Exercise 5.4 Class 12 Maths Chapter 5 Continuity and Differentiability, which makes them the ideal choice of 10,000+ students, who are going to appear in the final CBSE board examinations.

### Free Of Cost NCERT Solutions From Professional & Experienced Experts

Vedantu provides students like you with quality, and effective Exercise 5.4 notes absolutely free. You won’t be charged anything or even a single penny to have the complete access of our helpful NCERT Maths Textbook Solutions.

### An Opportunity To Get Ex 5.4 Class 12 Solutions From Top Academic Subject Experts

You will get to learn about one of the most crucial Continuity and Differentiability concepts from the top experts. While using Vedantu’s PDF notes, you get the chance to absorb the best information from the bets professionals themselves.

This way, you can get to learn every important concept and can ensure there is nothing left which is not clear in your mind. It is definitely going to help you a lot to put your best in the final board examinations.

### CBSE Class 12 Exercise 5.4 Solutions Followed Up According To The NCERT Guidelines

Vedantu clearly follows up with the NCERT guidelines in their PDF notes. Our team thoroughly understands the importance of class 12th in every student’s life and the necessity of following up with the NCERT guidelines your good & will. Class 12 Exercise 5.4 notes have all things and concepts which would help you to give your best in the final board examination.

### Learning NCERT Maths Class 12th Textbook Will Become Fun For You

Vedantu strongly believes and sticks to the proverb “learning should be fun instead of just cramming. This is something which is actually followed by our experts while preparing the notes and solutions.

Rather than just adding a point to point questions and answers of chapter 5 exercise 5.4, we add a complete and brief description of every question. Added tips, notes, illustrations, visuals, and diagrams are there to give you a better understanding of the complete 5.4 exercise.

### The Must Go-To-Option For Revising In Limited Time

Obviously, there comes a time when you may not have plenty of time to go through the complete exercise. In that case, Vedantu’s NCERT Ex 5.4 class 12 Maths solutions can be the ideal and perfect solution for you.

With these notes, you can quickly gain complete knowledge of Continuity and Differentiability concept. It is actually going to save a lot of your time and make you do well in your exams.

### Download Vedantu’s NCERT Solutions

Vedantu’s Class 12th Exercise 5.4 NCERT Solutions can provide you with the complete solutions of the entire exercise. You can download the PDF today from our website and get the full access to the answers.

Share this with your friends
SHARE
TWEET
SHARE
SUBSCRIBE