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NCERT Solutions

Class 12

Maths

Chapter 5 Continuity And Differentiability Exercise 5.4

NCERT Solutions for Class 12 Maths Chapter 5 Continuity And Differentiability Ex 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.

NCERT Solutions for Maths Class 12 Chapter 5 Exercise 5.4 - FREE PDF Download

NCERT Solutions for Exercise 5.4 Maths Class 12 Chapter 5 - Continuity and Differentiability contains all the problems in Exercise 5.4 completely and correctly. This exercise focuses on important ideas such as continuity and differentiability, as well as how they are applied to specific problems. The step-by-step solutions make it easy to learn about these complex problems.

Table of Content

It is important to pay attention to the methods used to demonstrate continuity and differentiation in NCERT Solutions for Class 12 Maths. Concentrate on knowing the definitions and theorems, as these are required to solve the questions correctly. Practising these solutions could help students improve their understanding and solve problems in the CBSE Class 12 Maths Syllabus.

Glance on NCERT Solutions Maths Chapter 5 Exercise 5.4 Class 12 | Vedantu

Exercise 5.4 Class 12 explains exponential and logarithmic functions, key concepts in calculus.
Exponential functions are of the form $f\left ( x \right )=a^{x}$, where a is a positive constant.
Logarithmic functions, the inverses of exponential functions, are expressed as $f\left ( x \right )=log_{a}\left ( x \right )$.
The exercise includes the differentiation of exponential functions, such as $e^{x}$ and $a^{x}$, which is essential for many applications.
It also covers the differentiation of logarithmic functions, like $ln\left ( x \right )$ and $log_{a}\left ( x \right )$, which frequently appear in calculus problems.
These functions are used to model growth and decay, which helps us understand natural and social changes.
There are 10 fully solved questions in Chapter 5 Exercise 5.4 Class 12 Continuity and Differentiability.

Formulas Used in Class 12 Chapter 5 Exercise 5.4

1. Derivative of Exponential Functions:

$\dfrac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x} \right )=e^{x}$
$\dfrac{\mathrm{d} }{\mathrm{d} x}\left ( a^{x} \right )=a^{x}ln(a)$

2. Derivative of Logarithmic Functions:

$\dfrac{\mathrm{d}}{\mathrm{d} x}(ln(x))=\dfrac{1}{x}$
$\dfrac{\mathrm{d}}{\mathrm{d} x}(log_{a}(x))=\dfrac{1}{x ln(a)}$

3. Chain Rule:

$\dfrac{\mathrm{d} }{\mathrm{d} x}[f(g(x))]={f}'(g(x))\cdot {g}'(x)$

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 for Class 12 Maths Chapter 5 Continuity And Differentiability Ex 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 for Class 12 Maths Chapter 5 Continuity And Differentiability Ex 5.4

NCERT Solutions for Maths Class 12 Chapter 5 Exercise 5.4 - FREE PDF Download

Glance on NCERT Solutions Maths Chapter 5 Exercise 5.4 Class 12 | Vedantu

Formulas Used in Class 12 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 for Class 12 Maths Chapter 5 Continuity And Differentiability Ex 5.4