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

NCERT Solutions

Class 12

Maths

Chapter 5 Continuity And Differentiability Exercise 5.4

Last updated date: 27th Jul 2024

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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

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

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

3. Formulas Used in Class 12 Chapter 5 Exercise 5.4

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

4.1

Exercise 5.4

5. Class 12 Maths Chapter 5: Exercises Breakdown

6. CBSE Class 12 Maths Chapter 5 Other Study Materials

7. Chapter-Specific NCERT Solutions for Class 12 Maths

FAQs

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

Exercise	Number of Questions
Exercise 5.1	34 Questions & Solutions (10 Short Answers, 24 Long Answers)
Exercise 5.2	10 Questions & Solutions (2 Short Answers, 8 Long Answers)
Exercise 5.3	15 Questions & Solutions (9 Short Answers, 6 Long Answers)
Exercise 5.5	18 Questions & Solutions (4 Short Answers, 14 Long Answers)
Exercise 5.6	11 Questions & Solutions
Exercise 5.7	17 Questions & Solutions (10 Short Answers, 7 Long Answers)
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 Important Formulas
4	Class 12 Continuity and Differentiability NCERT Exemplar Solution
5	Class 12 Continuity RD Sharma Solutions
6	Class 12 Differentiability RD Sharma Solutions
6	Class 12 Continuity and Differentiability RS Aggarwal Solutions

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

NCERT Solutions Class 12 Chapter-wise Maths PDF

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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

1. What is the main focus of Ex 5.4 Class 12 in NCERT Solutions for Maths Chapter 5?

The main focus of Ex 5.4 Class 12 is on understanding the differentiation of exponential and logarithmic functions. This includes applying the rules for differentiating these functions and solving related problems, which is crucial for mastering concepts in calculus.

2. How do you differentiate an exponential function in Ex 5.4 Class 12?

To differentiate an exponential function, you use the formula $\frac{\mathrm{d} }{\mathrm{d} x}(a^{x})=a^{x}ln\left ( a \right )$ for any base a. For the natural exponential function $e^{x}$, the derivative is $\frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x} \right )=e^{x}$.

3. What is the importance of logarithmic functions in Ex 5.4 Class 12?

Logarithmic functions are important in Ex 5.4 Class 12 because they help in simplifying complex expressions involving exponentials. Their differentiation rules, such as $\frac{\mathrm{d} }{\mathrm{d} x}\left ( ln\left ( x \right ) \right )=\frac{1}{x}$ and $\frac{\mathrm{d} }{\mathrm{d} x}\left ( log_{a}\left ( x \right ) \right )=\frac{1}{xln\left ( a \right )}$.

4. Why is it important to understand the properties of exponential functions in Chapter 5 Class 12 Maths Ex 5.4?

Understanding the properties of exponential functions is important because they frequently appear in various mathematical and real-world applications. Knowing how to differentiate and integrate these functions is fundamental for solving problems in calculus and other advanced topics.

5. How does Class 12 Maths Ex 5.4 help in preparing for board exams?

Class 12 Maths Ex 5.4 helps in preparing for board exams by providing practice in differentiating exponential and logarithmic functions, which are commonly tested concepts. Mastery of these topics enhances problem-solving skills and boosts confidence in tackling related questions in the exams.

6. Explain the significance of the natural logarithm function in calculus of Class 12 Maths Ex 5.4.

The natural logarithm function, ln(x), is significant in calculus because it simplifies the differentiation and integration of exponential functions. It also has important applications in solving growth and decay problems, making it a vital tool in various mathematical analyses.

7. What role does Class 12 Ex 5.4 play in understanding continuous functions?

Class 12 Ex 5.4 plays a role in understanding continuous functions by illustrating how exponential and logarithmic functions maintain continuity. These functions are continuous for all real numbers, and their differentiation properties help in analysing the behaviour of continuous functions in calculus.

8. How does Vedantu’s solution approach benefit students in mastering Class 12 Ex 5.4?

Vedantu’s solution approach benefits students by providing clear, step-by-step explanations for each problem in Class 12 Ex 5.4. This helps students grasp the underlying concepts of continuity and differentiability more effectively, reinforcing their understanding and improving their ability to solve similar problems independently.

9. Why is it important to practice Class 12 Ex 5.4 for calculus understanding?

Practising Class 12 Ex 5.4 is crucial for understanding calculus because it solidifies the concepts of continuity and differentiability. It allows students to apply their theoretical knowledge of exponential and logarithmic functions, enhancing their grasp of these fundamental topics in calculus.

10. How does working on Exercise 5.4 Class 12 Maths enhance analytical skills?

Working on Exercise 5.4 Class 12 Maths enhances analytical skills by requiring students to approach problems methodically and think critically. Detailed solutions help students learn how to break down complex problems into simpler steps, improving their ability to analyse and solve similar questions independently.

12. What are some real-world applications of the concepts covered in Exercise 5.4 Class 12 Maths?

Exponential and logarithmic functions are widely used in real-world applications such as modelling population growth, radioactive decay, and financial calculations like compound interest. These functions help students understand and solve practical problems related to growth and decay across various fields, for more refer Ex 5.4 Maths Class 12 page.

13. How do the NCERT Solutions for Exercise 5.4 Class 12 Maths ensure a thorough understanding of the topics?

The NCERT Solutions for Exercise 5.4 Class 12 Maths ensure thorough understanding by providing detailed, step-by-step explanations for each problem. This helps students grasp key concepts and methods, making it easier to understand and apply the material, and prepares them for more advanced calculus topics.