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

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NCERT Solutions for Class 12 Maths Chapter 5 (Ex 5.4)

Vedantu gives a whole new world to the way you learn and study in 12th Class. With the help of our Ex-5.4 Class 12 NCERT Solutions, you can learn faster, and in an effective way to get things done in the right direction. You can access our Free PDF of NCERT Solutions for Ex 5.4 Class 12 prepared by our professional teachers as per CBSE guidelines. This way, you can revise the entire exercise in a limited period of time and score higher marks in your upcoming examinations.


NCERT Solutions for Class 12


Class 12 Maths

Chapter Name:

Chapter 5 - Continuity and Differentiability


Exercise - 5.4


Text, Videos, Images and PDF Format

Academic Year:



English and Hindi

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Competitive Exams after 12th Science

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

Exercise 5.4

1. Differentiating the following w.r.t. \[x:\frac{{{e^x}}}{{\sin x}}\] 


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 ${e^{{{\sin }^{ - 1}}x}}$ 


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. $x:\sin ({\tan ^{ - 1}}{e^{ - x}})$ 


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. \[x:\log (\cos {e^x})\] 


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


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. $x:\sqrt {{e^{\sqrt x }}} ,x > 0$ 


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. $x:\log (\log x),x > 1$


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. $x:\frac{{\cos x}}{{\log x}},x > 0$


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. $x:\cos (\log x + {e^x}),x > 0$


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$ 

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