NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.4

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Chapter 5 Ex.5.4 Question 1

Differentiating the following wrt x: \(\frac{{{e^x}}}{{\sin x}}\)

Solution

Let \(y = \frac{{{e^x}}}{{\sin x}}\)

By using the quotient rule, we get

\[\begin{align}\frac{{dy}}{{dx}} &= \frac{{\sin x\frac{d}{{dx}}\left( {{e^x}} \right) - {e^x}\frac{d}{{dx}}\left( {\sin x} \right)}}{{{{\sin }^2}x}}\\& = \frac{{\sin x.\left( {{e^x}} \right) - {e^x}.\left( {\cos x} \right)}}{{{{\sin }^2}x}}\\ &= \frac{{{e^x}\left( {\sin x - \cos x} \right)}}{{{{\sin }^2}x}}\end{align}\]

Chapter 5 Ex.5.4 Question 2

Differentiating the following \({e^{{{\sin }^{ - 1}}x}}\)

Solution

Let \(y = {e^{{{\sin }^{ - 1}}x}}\)

By using the quotient rule, we get

\[\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{e^{{{\sin }^{ - 1}}x}}} \right)\\ &= {e^{{{\sin }^{ - 1}}x}}.\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)\\ &= {e^{{{\sin }^{ - 1}}x}}.\frac{1}{{\sqrt {1 - {x^2}} }}\\ &= \frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}\\ &= \frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }},x \in \left( { - 1,1} \right)\end{align}\]

Chapter 5 Ex.5.4 Question 3

Differentiating the following wrt \(x:{e^{{x^3}}}\)

Solution

Let \(y = {e^{{x^3}}}\)

By using the quotient rule, we get

\[\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right)\\ &= {e^{{x^3}}}.\frac{d}{{dx}}\left( {{x^3}} \right)\\& = {e^{{x^3}}}.3{x^2}\\ &= 3{x^2}{e^{{x^3}}}\end{align}\]

Chapter 5 Ex.5.4 Question 4

Differentiate the following wrt \(x:\sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\)

Solution

Let \(y = \sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\)

By using the chain rule, we get

\[\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {\sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)} \right]\\ &= \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right).\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\\& = \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right).\frac{1}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}}.\frac{d}{{dx}}\left( {{e^{ - x}}} \right)\\& = \frac{{\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{ - 2x}}}}.{e^{ - x}}.\frac{d}{{dx}}\left( { - x} \right)\\ &= \frac{{{e^{ - x}}\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{ - 2x}}}} \times \left( { - 1} \right)\\ &= \frac{{ - {e^{ - x}}\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{ - 2x}}}}\end{align}\]

Chapter 5 Ex.5.4 Question 5

Differentiate the following wrt \(x:\log \left( {\cos {e^x}} \right)\)

Solution

Let \(y = \log \left( {\cos {e^x}} \right)\)

By using the chain rule, we get

\[\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {\log \left( {\cos {e^x}} \right)} \right]\\&= \frac{1}{{\cos {e^x}}}.\frac{d}{{dx}}\left( {\cos {e^x}} \right)\\ &= \frac{1}{{\cos {e^x}}}.\left( { - \sin {e^x}} \right).\frac{d}{{dx}}\left( {{e^x}} \right)\\ &= \frac{{ - \sin {e^x}}}{{\cos {e^x}}}.{e^x}\\& = - {e^x}\tan {e^x},{e^x} \ne \left( {2n + 1} \right)\frac{\pi }{2},n \in {\bf{N}}\end{align}\]

Chapter 5 Ex.5.4 Question 6

Differentiate the following wrt \(x:{e^x} + {e^{{x^2}}} + \ldots + {e^{{x^5}}}\)

Solution

\(\frac{d}{{dx}}\left( {{e^x} + {e^{{x^2}}} + \ldots + {e^{{x^5}}}} \right)\)

Differentiating wrt x, we get

\[\begin{align}\frac{d}{{dx}}\left( {{e^x} + {e^{{x^2}}} + \ldots + {e^{{x^5}}}} \right) &= \frac{d}{{dx}}\left( {{e^x}} \right) + \frac{d}{{dx}}\left( {{e^{{x^2}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^4}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^5}}}} \right)\\ &= {e^x} + \left[ {{e^{{x^2}}} \times \frac{d}{{dx}}\left( {{x^2}} \right)} \right] + \left[ {{e^{{x^3}}} \times \frac{d}{{dx}}\left( {{x^3}} \right)} \right] \\& \qquad + \left[ {{e^{{x^4}}} \times \frac{d}{{dx}}\left( {{x^4}} \right)} \right] + \left[ {{e^{{x^5}}} \times \frac{d}{{dx}}\left( {{x^5}} \right)} \right]\\ &= {e^x} + \left( {{e^{{x^2}}} \times 2x} \right) + \left( {{e^{{x^3}}} \times 3{x^2}} \right) + \left( {{e^{{x^4}}} \times 4{x^3}} \right) + \left( {{e^{{x^5}}} \times 5{x^4}} \right)\\& = {e^x} + 2x{e^{{x^2}}} + 3{x^2}{e^{{x^3}}} + 4{x^3}{e^{{x^4}}} + 5{x^4}{e^{{x^5}}}\end{align}\]

Chapter 5 Ex.5.4 Question 7

Differentiating the following wrt \(x:\sqrt {{e^{\sqrt x }}} ,x > 0\)

Solution

Let \(y = \sqrt {{e^{\sqrt x }}} \)

Then, \({y^2} = {e^{\sqrt x }}\)

Differentiating wrt x, we get

\({y^2} = {e^{\sqrt x }}\)

\[\begin{align}&\frac{d}{{dx}}\left( {{y^2}} \right) = \frac{d}{{dx}}\left( {{e^{\sqrt x }}} \right)\\& \Rightarrow \;2y\frac{{dy}}{{dx}} = {e^{\sqrt x }}\frac{d}{{dx}}\left( {\sqrt x } \right)\\& \Rightarrow \;2y\frac{{dy}}{{dx}} = {e^{\sqrt x }}\frac{1}{2}.\frac{1}{{\sqrt x }}\\ &\Rightarrow \;\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4y\sqrt x }}\\& \Rightarrow \;\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4\sqrt {{e^{\sqrt x }}} \sqrt x }}\\& \Rightarrow \;\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4\sqrt {x{e^{\sqrt x }}} }},x > 0\end{align}\]

Chapter 5 Ex.5.4 Question 8

Differentiating the following wrt \(x:\log \left( {\log x} \right),x > 1\)

Solution

Let \(y = \log \left( {\log x} \right)\)

By using the chain rule, we get

\[\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left[ {\log \left( {\log x} \right)} \right]\\ &= \frac{1}{{\log x}}.\frac{d}{{dx}}\left( {\log x} \right)\\ &= \frac{1}{{\log x}}.\frac{1}{x}\\ &= \frac{1}{{x\log x}},x > 1\end{align}\]

Chapter 5 Ex.5.4 Question 9

Differentiating the following wrt \(x:\frac{{\cos x}}{{\log x}},x > 0\)

Solution

Let \(y = \frac{{\cos x}}{{\log x}}\)

By using the quotient rule, we get

\[\begin{align}\frac{{dy}}{{dx}}& = \frac{{\frac{d}{{dx}}\left( {\cos x} \right).\log x - \cos x.\frac{d}{{dx}}\left( {\log x} \right)}}{{{{\left( {\log x} \right)}^2}}}\\& = \frac{{ - \sin x\log x - \cos x.\frac{1}{x}}}{{{{\left( {\log x} \right)}^2}}}\\& = - \left[ {\frac{{x\log x.\sin x + \cos x}}{{x{{\left( {\log x} \right)}^2}}}} \right],x > 0\end{align}\]

Chapter 5 Ex.5.4 Question 10

Differentiate the following wrt \(x:\cos \left( {\log x + {e^x}} \right),x > 0\)

Solution

Let \(y = \cos \left( {\log x + {e^x}} \right)\)

By using the chain rule, we get

\[\begin{align}\frac{{dy}}{{dx}} &= - \sin \left[ {\log x + {e^x}} \right].\frac{d}{{dx}}\left( {\log x + {e^x}} \right)\\ &= - \sin \left( {\log x + {e^x}} \right).\left[ {\frac{d}{{dx}}\left( {\log x} \right) + \frac{d}{{dx}}\left( {{e^x}} \right)} \right]\\& = - \sin \left( {\log x + {e^x}} \right).\left( {\frac{1}{x} + {e^x}} \right)\\ &= - \left( {\frac{1}{x} + {e^x}} \right)\sin \left( {\log x + {e^x}} \right),x > 0\end{align}\]

  
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