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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 Ramraj Saini | Updated on Dec 03, 2023 05:00 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.4

NCERT Solutions for Exercise 5.4 Class 12 Maths Chapter 5 Continuity and Differentiability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. You have already learned about the exponential and logarithmic functions in the previous classes. In the NCERT solutions for Class 12 Maths chapter 5 exercise 5.4, you will learn about the differentiation and logarithmic functions. The differentiation of simple exponential and logarithmic functions is very easy. In exercise 5.4 Class 12 Maths you will get questions related to the differentiation of composite functions containing the exponential and logarithmic functions.

As you learned the differentiation of composite functions in Class 12 Maths NCERT book exercise 5.2 already, you can use that concept to solve Class 12 Maths ch 5 ex 5.4 problems. Don't confuse with logarithmic differentiation, it is different from the differentiation of logarithmic function. You will about the logarithmic differentiation in the next exercise of Class 12 Maths NCERT syllabus. 12th class Maths exercise 5.4 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4

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Continuity and Differentiability Exercise: 5.4

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

\frac{e ^x }{\sin x }

Answer:

Given function is
f(x)=\frac{e ^x }{\sin x }
We differentiate with the help of Quotient rule
f^{'}(x)=\frac{\frac{d(e^x)}{dx}.\sin x-e^x.\frac{(\sin x)}{dx} }{\sin^2 x }
=\frac{e^x.\sin x-e^x.\cos }{\sin^2 x } = \frac{e^x(\sin x-\cos x)}{\sin^2x}
Therefore, the answer is \frac{e^x(\sin x-\cos x)}{\sin^2x}

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

e ^{\sin ^{-1}x}

Answer:

Given function is
f(x)=e ^{\sin ^{-1}x}
Let g(x)={\sin ^{-1}x}
Then,
f(x)=e^{g(x)}
Now, differentiation w.r.t. x
f^{'}(x)=g^{'}(x).e^{g(x)} -(i)
g(x) = \sin^{-1}x \Rightarrow g^{'}(x ) = \frac{1}{\sqrt{1-x^2}}
Put this value in our equation (i)
f^{'}(x) = \frac{1}{\sqrt{1-x^2}}.e^{\sin^{-1}x} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}

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

e ^x^{ ^3}

Answer:

Given function is
f(x)=e ^{x^3}
Let g(x)=x^3
Then,
f(x)=e^{g(x)}
Now, differentiation w.r.t. x
f^{'}(x)=g^{'}(x).e^{g(x)} -(i)
g(x) = x^3 \Rightarrow g^{'}(x ) =3x^2
Put this value in our equation (i)
f^{'}(x) =3x^2.e^{x^3}
Therefore, the answer is 3x^2.e^{x^3}

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}\\\Rightarrow g^{'}(x) = \frac{d(\tan^{-1}e^{-x})}{dx}.\frac{d(e^{-x})}{dx}= \frac{1}{1+(e^{-x})^2}.-e^{-x} = \frac{-e^{-x}}{1+e^{-2x}}
Put this value in our equation (i)
f^{'}(x) =\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})
Therefore, the answer is \frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})

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

\log (\cos e ^x )

Answer:

Given function is
f(x)=\log (\cos e ^x )
Let's take g(x ) = \cos e^{x}
Now, our function reduces to
f(x) = \log(g(x))
Now,
f^{'}(x) = g^{'}(x).\frac{1}{g(x)} -(i)
And
g(x)=\cos e^{x}\\\Rightarrow g^{'}(x) = \frac{d(\cos e^{x})}{dx}.\frac{d(e^{x})}{dx}= (-\sin e^x).e^{x} = -e^x.\sin e^x
Put this value in our equation (i)
f^{'}(x) =-e^x.\sin e^x.\frac{1}{\cos e^x} = -e^x.\tan e^x \ \ \ \ \ (\because \frac{\sin x}{\cos x}=\tan x)
Therefore, the answer is -e^x.\tan e^x,\ \ \ e^x\neq (2n+1)\frac{\pi}{2},\ \ n\in N

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

e ^x + e ^{x^2} + .....e ^{x^5}

Answer:

Given function is
f(x)= e ^x + e ^{x^2} + .....e ^{x^5}
Now, differentiation w.r.t. x is
f^{'}(x)= \frac{d(e^x)}{dx}.\frac{d(x)}{dx}+\frac{d(e^{x^2})}{dx}.\frac{d(x^2)}{dx}+\frac{d(e^{x^3})}{dx}.\frac{d(x^3)}{dx}+\frac{d(e^{x^4})}{dx}.\frac{d(x^4)}{dx}+\frac{d(e^{x^5})}{dx}.\frac{d(x^5)}{dx}
=e^x.1+e^{x^2}.2x+e^{x^3}.3x^2+e^{x^4}.4x^3+e^{x^5}.5x^4
=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}
Therefore, answer is e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}

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

\sqrt { e ^{ \sqrt x }} , x > 0

Answer:

Given function is
f(x)=\sqrt { e ^{ \sqrt x }}
Lets take g(x ) = \sqrt x
Now, our function reduces to
f(x) = \sqrt {e^{g(x)}}
Now,
f^{'}(x) = g^{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.\frac{d({e^{g(x)}})}{dx} = g{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.{e^{g(x)}} = \frac{g^{'}(x).e^{g(x)}}{2.\sqrt{e^{g(x)}}} = \frac{g^{'}(x).e^{\sqrt x}}{2.\sqrt{e^{\sqrt x}}} -(i)
And
g(x)=\sqrt x\\\Rightarrow g^{'}(x) = \frac{(\sqrt x)}{dx}=\frac{1}{2\sqrt x}
Put this value in our equation (i)
f^{'}(x) =\frac{e^{\sqrt x}}{2\sqrt x.2.\sqrt{e^{\sqrt x}}} = \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}
Therefore, the answer is \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}.\ \ 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).\frac{1}{g(x)} -(i)
And
g(x)=\log x\\\Rightarrow g^{'}(x) = \frac{1}{x}
Put this value in our equation (i)
f^{'}(x) =\frac{1}{x}.\frac{1}{\log x} = \frac{1}{x\log x}
Therefore, the answer is \frac{1}{x\log x}, \ \ x>1

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

\frac{\cos x }{\log x} , x > 0

Answer:

Given function is
f(x)=\frac{\cos x }{\log x}
We differentiate with the help of Quotient rule
f^{'}(x)=\frac{\frac{d(\cos x)}{dx}.\log x-\cos x.\frac{(\log x)}{dx} }{(\log x)^2 }
=\frac{(-\sin x).\log x-\cos x.\frac{1}{x} }{(\log x)^2 } = \frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}
Therefore, the answer is \frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}

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

\cos ( log x + e ^x ) , x > 0

Answer:

Given function is
f(x)=\cos ( log x + e ^x )
Lets take g(x) = ( log x + e ^x )
Then , our function reduces to
f(x) = \cos (g(x))
Now, differentiation w.r.t. x is
f^{'}(x) = g^{'}(x)\(-\sin) (g(x)) -(i)
And
g(x) = ( log x + e ^x )
g^{'}(x)= \frac{d(\log x)}{dx}+\frac{d(e^x)}{dx} = \frac{1}{x}+e^x
Put this value in our equation (i)
f^{'}(x) = -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x)
Therefore, the answer is -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x), x>0

More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4:-

Class 12th Maths chapter 5 exercise 5.4 is all about the differentiation of composite functions containing exponential and logarithmic functions. There are two examples and few theorems given before the Class 12 Maths ch 5 ex 5.4 are related to that only. There are 10 questions in this exercise related to finding differentiation of exponential and logarithmic functions. You can easily solve these problems if you have done good practice of chain rule problems in exercise 5.2 of this chapter.

Also Read| Continuity and Differentiability Class 12th Chapter 5 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4:-

  • The differentiation of logarithmic and exponential is an important concept that will be useful not only in this maths but in all the other subjects also.
  • NCERT solutions for Class 12 Maths chapter 5 exercise 5.4 are useful in learning this important concept of differentiation.
  • Class 12 Maths chapter 5 exercise 5.4 solutions are designed following the guideline of CBSE.
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Key Features Of NCERT Solutions for Exercise 5.4 Class 12 Maths Chapter 5

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 5.4 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 5.4, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 5.4 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 5.4 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 5.4 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 5.4 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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Frequently Asked Question (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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i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

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Divyan 06 August,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,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

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i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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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.

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Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

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If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

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kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
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  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

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