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NCERT Solutions for Exercise 5.4 Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solutions for Exercise 5.4 Class 12 Maths Chapter 5 - Continuity and Differentiability

Edited By Komal Miglani | Updated on Apr 23, 2025 11:09 PM IST | #CBSE Class 12th

If continuity is like playing a song without any pauses, then differentiability means that there are no high pitches and sudden jumps in tone, just silky smooth transitions. Finding the derivatives of exponential and logarithmic functions is an important aspect of calculus. In exercise 5.4 of the chapter Continuity and Differentiability, we explore different methods to easily differentiate functions involving exponential and logarithmic functions. This article on the NCERT Solutions for exercise 5.4, class 12 maths chapter 5 - Continuity and Differentiability, offers an easy-to-understand and step-by-step solution for the problems present in the exercise, so that students will get a better understanding of the methods and logic. For syllabus, notes, and PDF, refer to this link: NCERT

LiveCBSE Result 2025 (LIVE): Class 10, 12 marksheet likely today at cbse.gov.in, Digilocker; websites to checkMay 12, 2025 | 10:32 PM IST

No, the Central Board of Secondary Education is yet to declare CBSE Class 10 results 2025. Once declared, the CBSE 10th result link 2025 will be available on the official website, results.cbse.nic.in.

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Class 12 Maths Chapter 5 Exercise 5.4 Solutions: Download PDF

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Continuity and Differentiability Exercise: 5.4

Question:1. Differentiate the following w.r.t. x:

exsin⁡x

Answer:

Given function is
f(x)=exsin⁡x
We differentiate with the help of Quotient rule
f′(x)=d(ex)dx⋅sin⁡x−ex⋅d(sin⁡x)dxsin2⁡x=ex⋅sin⁡x−ex⋅cos⁡xsin2⁡x=ex(sin⁡x−cos⁡x)sin2⁡x
Therefore, the answer is ex(sin⁡x−cos⁡x)sin2⁡x

Question:2. Differentiate the following w.r.t. x:

esin−1⁡x

Answer:

Given function is
f(x)=esin−1⁡x
Let g(x)=sin−1⁡x
Then,
f(x)=eg(x)
Now, differentiation w.r.t. x
f′(x)=g′(x).eg(x) -(i)
g(x)=sin−1⁡x⇒g′(x)=11−x2
Put this value in our equation (i)
f′(x)=11−x2.esin−1⁡x=esin−1⁡x1−x2

Question:3. Differentiate the following w.r.t. x:

ex3

Answer:

Given function is
f(x)=ex3
Let g(x)=x3
Then,
f(x)=eg(x)
Now, differentiation w.r.t. x
f′(x)=g′(x).eg(x) -(i)

g(x)=x3⇒g′(x)=3x2

Put this value in our equation (i)
f′(x)=3x2.ex3
Therefore, the answer is 3x2.ex3

Question:4. Differentiate the following w.r.t. x:

sin⁡(tan−1⁡e−x)

Answer:

Given function is
f(x)=sin⁡(tan−1⁡e−x)
Let's take g(x)=tan−1⁡e−x
Now, our function reduces to
f(x)=sin⁡(g(x))
Now,
f′(x)=g′(x)cos⁡(g(x)) -(i)
And
g(x)=tan−1⁡e−x⇒g′(x)=d(tan−1⁡e−x)dx.d(e−x)dx=11+(e−x)2.−e−x=−e−x1+e−2x
Put this value in our equation (i)
f′(x)=−e−x1+e−2xcos⁡(tan−1⁡e−x)
Therefore, the answer is −e−x1+e−2xcos⁡(tan−1⁡e−x)

Question:5. Differentiate the following w.r.t. x:

log⁡(cos⁡ex)

Answer:

Given function is
f(x)=log⁡(cos⁡ex)
Let's take g(x)=cos⁡ex
Now, our function reduces to
f(x)=log⁡(g(x))
Now,
f′(x)=g′(x).1g(x) -(i)
And
g(x)=cos⁡ex⇒g′(x)=d(cos⁡ex)dx.d(ex)dx=(−sin⁡ex).ex=−ex.sin⁡ex
Put this value in our equation (i)
f′(x)=−ex.sin⁡ex.1cos⁡ex=−ex.tan⁡ex     (∵sin⁡xcos⁡x=tan⁡x)
Therefore, the answer is −ex.tan⁡ex,   ex≠(2n+1)π2,  n∈N

Question:6. Differentiate the following w.r.t. x:

ex+ex2+.....ex5

Answer:

Given function is
f(x)=ex+ex2+.....ex5
Now, differentiation w.r.t. x is
f′(x)=d(ex)dx.d(x)dx+d(ex2)dx.d(x2)dx+d(ex3)dx.d(x3)dx+d(ex4)dx.d(x4)dx+d(ex5)dx.d(x5)dx
=ex.1+ex2.2x+ex3.3x2+ex4.4x3+ex5.5x4
=ex+2xex2+3x2ex3+4x3ex4+5x4ex5
Therefore, answer is ex+2xex2+3x2ex3+4x3ex4+5x4ex5

Question:7. Differentiate the following w.r.t. x:

ex,x>0

Answer:

Given function is
f(x)=ex
Lets take g(x)=x
Now, our function reduces to
f(x)=eg(x)
Now,
f′(x)=g′(x)⋅12eg(x)⋅d(eg(x))dx=g′(x)⋅12eg(x)⋅eg(x)=g′(x)⋅eg(x)2eg(x)=g′(x)⋅ex2ex -(i)
And
g(x)=x⇒g′(x)=(x)dx=12x
Put this value in our equation (i)
f′(x)=ex2x⋅2ex=ex4xex
Therefore, the answer is ex4xex.  x>0

Question:8 Differentiate the following w.r.t. x: log⁡(log⁡x),x>1

Answer:

Given function is
f(x)=log⁡(log⁡x)
Lets take g(x)=log⁡x
Now, our function reduces to
f(x)=log⁡(g(x))
Now,
f′(x)=g′(x).1g(x) -(i)
And
g(x)=log⁡x⇒g′(x)=1x
Put this value in our equation (i)
f′(x)=1x.1log⁡x=1xlog⁡x
Therefore, the answer is 1xlog⁡x,  x>1

Question:9. Differentiate the following w.r.t. x:

cos⁡xlog⁡x,x>0

Answer:

Given function is
f(x)=cos⁡xlog⁡x
We differentiate with the help of Quotient rule
f′(x)=d(cos⁡x)dx.log⁡x−cos⁡x.(log⁡x)dx(log⁡x)2
=(−sin⁡x).log⁡x−cos⁡x.1x(log⁡x)2=−(xsin⁡xlog⁡x+cos⁡x)x(log⁡x)2
Therefore, the answer is −(xsin⁡xlog⁡x+cos⁡x)x(log⁡x)2

Question:10. Differentiate the following w.r.t. x:

cos⁡(logx+ex),x>0

Answer:

Given function is
f(x)=cos⁡(logx+ex)
Lets take g(x)=(logx+ex)
Then , our function reduces to
f(x)=cos⁡(g(x))
Now, differentiation w.r.t. x is
f′(x)=g′(x)(−sin⁡(g(x)))
And
g(x)=(logx+ex)
g′(x)=d(log⁡x)dx+d(ex)dx=1x+ex
Put this value in our equation (i)
f′(x)=−(1x+ex)sin⁡(log⁡x+ex)
Therefore, the answer is −(1x+ex)sin⁡(log⁡x+ex),x>0


Also Read,

Topics covered in Chapter 5, Continuity and Differentiability: Exercise 5.4

The main topics covered in Chapter 5 of continuity and differentiability, exercises 5.4 are:

  • Derivatives of Exponential functions: Differentiation of exponential functions is very basic in calculus. In particular derivative of ex is very simple.
    ddx(ex)=ex
    Also, for the general exponent function, the derivative is as follows:
    ddx(ax)=axln⁡a
  • Derivatives of Logarithmic functions: Logarithmic functions help in simplifying many complex calculations. The basic differentiation rule of a logarithmic function is:
    ddx(ln⁡x)=1x
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Frequently Asked Questions (FAQs)

1. What is the derivative of e^x ?

The derivative of e^x is e^x.

2. What is the derivative of log(x) ?

The derivative of log(x) is 1/x.

3. what is the derivative of sin(x) ?

The derivative of sin(x) is cos(x).

4. what is the derivative of cos(x) ?

The derivative of cos(x) is -sin(x).

5. What is the derivative of sin(x^2) ?

The derivative of sin(x^2) is 2x cos(x^2).

6. Which is the best book for Maths CBSE Class 12 ?

NCERT textbook is the best book for CBSE board exams which you should follow. You don't need any additional books for the CBSE board exams.

7. How may questions are there in the exercise 5.4 Class 12 Maths ?

A total of 10 questions are there in exercise 5.4 CBSE Class 12 Maths. For more questions students can refer to the NCERT exemplar questions.

8. Find the derivative of tan (x) ?

The derivative of tan (x) is sec^2(x).

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Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

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Manasvini Gupta 30 January,2025
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

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Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

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Jefferson 25 September,2024
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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
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