Class 12 ncert solution math exercise 5.4

    EXERCISE 5.4 ( Differentiation )

Differentiating the following w.r.t.(Class 12 ncert solution math exercise 5.4)

Question 1: \frac{e^x}{\sin x}.

Solution:Let y=\frac{e^x}{\sin x}

By using the quotient rule, we get

\frac{dy}{dx} =\frac{\sin x \frac{d}{d x}\left(e^x\right)-e^x \frac{d}{d x}(\sin x)}{\sin ^2 x}

=\frac{\sin x \cdot\left(e^x\right)-e^x \cdot(\cos x)}{\sin ^2 x}

=\frac{e^x(\sin x-\cos x)}{\sin ^2 x}

Question 2: e^{\sin ^{-1} x}

Solution:Let y=e^{\sin ^{-1} x}

By using the quotient rule, we get

\frac{d y}{d x} =\frac{d}{d x}\left(e^{\sin ^{-1} x}\right)

=e^{\sin ^{-1} x} \cdot \frac{d}{d x}\left(\sin ^{-1} x\right)

=e^{\sin ^{-1} x} \cdot \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(-1,1)

Question 3: e^{x^3}

Solution:Let y=e^{x^3}

By using the quotient rule, we get

\frac{d y}{d x} =\frac{d}{d x}\left(e^{x^3}\right)

=e^{x^3} \cdot \frac{d}{d x}\left(x^3\right)

=e^{x^3} \cdot 3 x^2

=3 x^2 e^{x^3}

Question 4:  \sin \left(\tan ^{-1} e^{-x}\right)

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

By using the chain rule, we get

\frac{d y}{d x} =\frac{d}{d x}\left[\sin \left(\tan ^{-1} e^{-x}\right)\right]

=\cos \left(\tan ^{-1} e^{-x}\right) \cdot \frac{d}{d x}\left(\tan ^{-1} e^{-x}\right)

=\cos \left(\tan ^{-1} e^{-x}\right) \cdot \frac{1}{1+\left(e^{-x}\right)^2} \cdot \frac{d}{d x}\left(e^{-x}\right)

=\frac{\cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}} \cdot e^{-x} \cdot \frac{d}{d x}(-x)

=\frac{e^{-x} \cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}} \times(-1)

=\frac{-e^{-x} \cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}}

Question 5: \log \left(\cos e^x\right)

Solution:Let y=\log \left(\cos e^x\right)

By using the chain rule, we get

\frac{d y}{d x} =\frac{d}{d x}\left[\log \left(\cos e^x\right)\right]

=\frac{1}{\cos e^x} \cdot \frac{d}{d x}\left(\cos e^x\right)

=\frac{1}{\cos e^x} \cdot\left(-\sin e^x\right) \cdot \frac{d}{d x}\left(e^x\right)

=\frac{-\sin e^x}{\cos e^x} \cdot e^x

=-e^x \tan e^x, e^x \neq(2 n+1) \frac{\pi}{2}, n \in \mathbf{N}

Question 6: e^x+e^{x^2}+\ldots+e^{x^5}

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

Differentiating wrt x, we get

\frac{d}{d x}\left(e^x+e^{x^2}+\ldots+e^{x^5}\right) =\frac{d}{d x}\left(e^x\right)+\frac{d}{dx}\left(e^{x^2}\right)+\frac{d}{d x}\left(e^{x^3}\right)+\frac{d}{d x}\left(e^{x^4}\right)+\frac{d}{d x}\left(e^{x^5}\right)

=e^x+\left[e^{x^2} \times \frac{d}{d x}\left(x^2\right)\right]+\left[e^{x^3} \times \frac{d}{d x}\left(x^3\right)\right]+\left[e^{x^4} \times \frac{d}{d x}\left(x^4\right)\right]+\left[e^{x^5} \times \frac{d}{d x}\left(x^5\right)\right]

=e^x+\left(e^{x^2} \times 2 x\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+2 x e^{x^2}+3 x^2 e^{x^3}+4 x^3 e^{x^4}+5 x^4 e^{x^5}

Question 7: \sqrt{e^{\sqrt{x}}}, x>0

Solution: Let y=\sqrt{e^{\sqrt{x}}}

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

Differentiating wrt x, we get


\frac{d}{d x}\left(y^2\right)=\frac{d}{d x}\left(e^{\sqrt{x}}\right)

\Rightarrow 2 y \frac{dy}{dx}=e^{\sqrt{x}} \frac{d}{d x}(\sqrt{x})

\Rightarrow 2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{1}{2} \cdot \frac{1}{\sqrt{x}}

\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 y \sqrt{x}}

\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}}

\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}, x>0

Question 8: \log (\log x), x>1

Solution: Let y=\log (\log x)

By using the chain rule, we get

\frac{d y}{d x} =\frac{d}{d x}[\log (\log x)]

=\frac{1}{\log x} \cdot \frac{d}{d x}(\log x)

=\frac{1}{\log x} \cdot \frac{1}{x}

=\frac{1}{x \log x}, x>1

Question 9:  \frac{\cos x}{\log x}, x>0

Solution: Let y=\frac{\cos x}{\log x}

By using the quotient rule, we get

\frac{d y}{d x} =\frac{\frac{d}{d x}(\cos x) \cdot \log x-\cos x \cdot \frac{d}{d x}(\log x)}{(\log x)^2}

=\frac{-\sin x \log x-\cos x \cdot \frac{1}{x}}{(\log x)^2}

=-\left[\frac{x \log x \cdot \sin x+\cos x}{x(\log x)^2}\right], x>0

Question 10: \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

\frac{d y}{d x} =-\sin \left[\log x+e^x\right] \cdot \frac{d}{d x}\left(\log x+e^x\right)

=-\sin \left(\log x+e^x\right) \cdot\left[\frac{d}{d x}(\log x)+\frac{d}{d x}\left(e^x\right)\right].

=-\sin \left(\log x+e^x\right) \cdot\left(\frac{1}{x}+e^x\right)

=-\left(\frac{1}{x}+e^x\right) \sin \left(\log x+e^x\right), x>0


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