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NCERT Solutions

Class 12

Maths

Chapter 5 Continuity And Differentiability Exercise 5.4

Ncert Solutions Class 12 Maths Chapter 5 Exercise 5 4

Q: 2. How does Exercise 5.4 strengthen understanding of continuity and differentiability for Class 12 students?

Exercise 5.4 provides practice in differentiating exponential and logarithmic functions, helping students internalise how these functions behave and maintain continuity. This deepens conceptual clarity for calculus questions in CBSE board exams.

Q: 3. What formulas are commonly required to solve problems in Class 12 Maths Chapter 5 Exercise 5.4?

The key formulas include:Derivative of exponential functions: $\frac{d}{dx} a^{x} = a^{x} \ln a$Derivative of natural exponent: $\frac{d}{dx} e^{x} = e^{x}$Derivative of logarithms: $\frac{d}{dx} \ln x = \frac{1}{x}$, $\frac{d}{dx} \log_{a} x = \frac{1}{x \ln a}$Chain rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

Q: 4. Why is it necessary to understand the properties of exponential and logarithmic functions before attempting Exercise 5.4 questions?

Understanding these properties is essential, as exponential and logarithmic functions have unique characteristics that affect their continuity, differentiability, and real-world applications. Familiarity with their behaviour ensures accurate application of derivatives and helps avoid common calculation errors.

Q: 8. What is the role of the chain rule in solving problems from NCERT Class 12 Maths Exercise 5.4?

The chain rule is critical for differentiating composite functions, such as $\ln(\cos e^{x})$ or $e^{g(x)}$. It ensures correct calculation when a function is applied to another function, which frequently appears in Exercise 5.4.

Q: 10. What is the importance of checking function domains when solving Class 12 Exercise 5.4 questions?

Checking domains avoids undefined values, such as negatives inside logarithms or divisions by zero. This ensures mathematical validity of each step and matches CBSE marking schemes, which penalise domain errors in answers.

An Overview of Ncert Solutions Class 12 Maths Chapter 5 Exercise 5 4

In NCERT Solutions Class 12 Maths Chapter 5 Exercise 5 4, you’ll learn how to handle exponential and logarithmic functions—important ideas if you want to master calculus and score well in board exams. This part of the chapter explains how to differentiate these functions, making concepts like growth and decay simple to understand. Vedantu’s step-by-step answers make even the trickiest sums easier to solve.

Table of Content

If you get stuck on continuity or the chain rule, these Class 12 Maths NCERT Solutions are a real lifesaver. You can also download full PDFs for free, so revision gets quick and hassle-free. For a complete guide to your CBSE Class 12 Maths syllabus, see the Class 12 Maths syllabus.

This chapter carries 9 marks in your CBSE exam, so practising from the NCERT Solutions for Class 12 Maths will definitely help you boost your score and clear your doubts faster.

Access NCERT Solutions for Maths Class 12 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$

Conclusion

Vedantu's NCERT Solutions for Maths Exercise 5.4 Class 12 Chapter 5 - Continuity and Differentiability provide clear answers to exponential and logarithmic functions. Focus on the differentiation concepts and how these functions are applied in real-world situations. Understanding these concepts is important for solving calculus problems. Practising Exercise 5.4 Class 12 will improve your problem solving skills and help you score well in exams.

Class 12 Maths Chapter 5: Exercises Breakdown

S.No.	Chapter 5 - Continuity and Differentiability Exercises in PDF Format
1	Class 12 Maths Chapter 5 Exercise 5.1 - 34 Questions & Solutions (10 Short Answers, 24 Long Answers)
2	Class 12 Maths Chapter 5 Exercise 5.2 - 10 Questions & Solutions (2 Short Answers, 8 Long Answers)
3	Class 12 Maths Chapter 5 Exercise 5.3 - 15 Questions & Solutions (9 Short Answers, 6 Long Answers)
4	Class 12 Maths Chapter 5 Exercise 5.5 - 18 Questions & Solutions (4 Short Answers, 14 Long Answers)
5	Class 12 Maths Chapter 5 Exercise 5.6 - 11 Questions & Solutions (7 Short Answers, 4 Long Answers)
6	Class 12 Maths Chapter 5 Exercise 5.7 - 17 Questions & Solutions (10 Short Answers, 7 Long Answers)
7	Class 12 Maths Chapter 5 Miscellaneous Exercise - 22 Questions & Solutions

CBSE Class 12 Maths Chapter 5 Other Study Materials

S.No.	Important Links for Chapter 5 Continuity and Differentiability
1	Class 12 Continuity and Differentiability Important Questions
2	Class 12 Continuity and Differentiability Revision Notes
3	Class 12 Continuity and Differentiability Formulas
4	Class 12 Continuity and Differentiability NCERT Exemplar Solutions
5	Class 12 Continuity and Differentiability RS Aggarwal Solutions

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

S.No.	NCERT Solutions Class 12 Maths Chapter-wise List
1	Chapter 1 - Relations and Functions Solutions
2	Chapter 2 - Inverse Trigonometric Functions Solutions
3	Chapter 3 - Matrices Solutions
4	Chapter 4 - Determinants Solutions
5	Chapter 5 - Continuity and Differentiability Solutions
6	Chapter 6 - Application of Derivatives Solutions
7	Chapter 7 - Integrals Solutions
8	Chapter 8 - Application of Integrals Solutions
9	Chapter 9 - Differential Equations Solutions
10	Chapter 10 - Vector Algebra Solutions
11	Chapter 11 - Three Dimensional Geometry Solutions
12	Chapter 12 - Linear Programming Solutions
13	Chapter 13 - Probability Solutions

Important Related Links for NCERT Class 12 Maths

S.No	Important Resources Links for Class 12 Maths
1	CBSE Class 12 Maths Revision Notes
2	CBSE Syllabus for Class 12 Maths
3	CBSE Class 12 Maths Important Questions
4	CBSE Class 12 Maths Previous Year Question Paper
5	CBSE Class 12 Maths Sample Paper
6	NCERT Books for Class 12 Maths
7	NCERT Exemplar for Class 12 Maths
8	CBSE Class 12 Maths Formulas
9	RS Aggarwal Class 12 Maths Solutions
10	RD Sharma Class 12 Maths Solutions

FAQs on Ncert Solutions Class 12 Maths Chapter 5 Exercise 5 4

1. What are the main steps involved in solving questions from NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4 as per CBSE 2025–26 syllabus?

To solve questions from Exercise 5.4, start with identifying the type of function (exponential, logarithmic), use the relevant derivative formula, apply the chain rule or quotient rule if needed, simplify the result, and check if the answer matches standard forms as demonstrated in NCERT Solutions for Class 12 Maths.

2. How does Exercise 5.4 strengthen understanding of continuity and differentiability for Class 12 students?

Exercise 5.4 provides practice in differentiating exponential and logarithmic functions, helping students internalise how these functions behave and maintain continuity. This deepens conceptual clarity for calculus questions in CBSE board exams.

3. What formulas are commonly required to solve problems in Class 12 Maths Chapter 5 Exercise 5.4?

The key formulas include:

Derivative of exponential functions: $\frac{d}{dx} a^{x} = a^{x} \ln a$
Derivative of natural exponent: $\frac{d}{dx} e^{x} = e^{x}$
Derivative of logarithms: $\frac{d}{dx} \ln x = \frac{1}{x}$, $\frac{d}{dx} \log_{a} x = \frac{1}{x \ln a}$
Chain rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

4. Why is it necessary to understand the properties of exponential and logarithmic functions before attempting Exercise 5.4 questions?

Understanding these properties is essential, as exponential and logarithmic functions have unique characteristics that affect their continuity, differentiability, and real-world applications. Familiarity with their behaviour ensures accurate application of derivatives and helps avoid common calculation errors.

5. How does step-by-step problem solving in NCERT Solutions for Exercise 5.4 help students avoid common mistakes?

Stepwise solutions break down each problem into logical stages, ensuring every formula and rule—like the chain rule—is applied correctly. This method minimises skipping steps and reduces the risk of missing critical calculation points, strengthening CBSE exam performance.

6. What misconceptions should students watch for when differentiating logarithmic or exponential expressions in Class 12?

Major misconceptions include confusing the base of the logarithm, forgetting the multiplication by $\ln a$ in $a^{x}$ derivatives, and misapplying the chain rule. Being attentive to function domains and composite functions is crucial to accurate problem solving.

7. In what ways are the concepts covered in Class 12 Maths Chapter 5 Exercise 5.4 used in real-life scenarios?

Exponential and logarithmic functions are pivotal in modelling population growth, radioactive decay, compound interest, and even data analysis. Learning their differentiation helps students apply mathematical techniques to scientific, economic, and engineering problems.

8. What is the role of the chain rule in solving problems from NCERT Class 12 Maths Exercise 5.4?

The chain rule is critical for differentiating composite functions, such as $\ln(\cos e^{x})$ or $e^{g(x)}$. It ensures correct calculation when a function is applied to another function, which frequently appears in Exercise 5.4.

9. How can practising the solved solutions from Exercise 5.4 improve analytical skills for board exam questions?

By working through detailed solutions, students enhance their ability to break down complex problems, identify correct methods, and approach unfamiliar questions logically—skills vital for tackling high-difficulty CBSE board problems.

10. What is the importance of checking function domains when solving Class 12 Exercise 5.4 questions?

Checking domains avoids undefined values, such as negatives inside logarithms or divisions by zero. This ensures mathematical validity of each step and matches CBSE marking schemes, which penalise domain errors in answers.

11. How does applying NCERT Solutions for Exercise 5.4 help in mastering advanced calculus in higher education?

Practising these solutions builds a strong foundation in differentiation of advanced functions, essential for topics like integration, differential equations, and mathematical modelling introduced in higher studies and competitive exams.

12. What methods are recommended for checking the correctness of your differentiated answers in Exercise 5.4?

To confirm correctness, re-calculate stepwise, substitute values into original and derived functions for verification, and cross-check with examples from NCERT Solutions for Class 12 Maths Chapter 5 as per CBSE 2025–26 guidelines.

S.No.	Related NCERT Solutions for Class 12 Maths
1	NCERT Book for Class 12 Maths in Hindi
2	NCERT Solutions for Class 12 Maths in hindi

Ncert Solutions Class 12 Maths Chapter 5 Exercise 5 4

An Overview of Ncert Solutions Class 12 Maths Chapter 5 Exercise 5 4

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

Exercise 5.4

Conclusion

Class 12 Maths Chapter 5: Exercises Breakdown

CBSE Class 12 Maths Chapter 5 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Related Links for NCERT Class 12 Maths in Hindi

Important Related Links for NCERT Class 12 Maths

FAQs on Ncert Solutions Class 12 Maths Chapter 5 Exercise 5 4