NCERT Solutions for Class 12 Maths Chapter 5 (Ex 5.4)
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Exercise:  Exercise  5.4 
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Access NCERT Solutions for Class 12 Maths Chapter5 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$
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