Differentiate the function ( x)^x + ^ –1 x w.r.t. x. - Vidyadip

Question

Differentiate the function $(\sin x)^x + \sin^{–1} \sqrt{x} w.r.t. x.$

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Answer

Given function is : $(\sin x)^{x}+\sin ^{-1} \sqrt{x}$
Let $y=(\sin x)^{x}+\sin ^{-1} \sqrt{x}$
Let $y = u + v$
$\Rightarrow u = (\sin x)^x$ and $\mathrm{v}=\sin ^{-1} \sqrt{\mathrm{x}}$
For, $\mathrm{u}=(\sin \mathrm{x})^{\mathrm{x}}$
Taking $\log$ on both sides, we get
$\log u = \log(\sin x)^x$
$\Rightarrow \log u = x.\log \sin x$
Now, differentiate both sides with respect to $x$
$\frac{d}{d x}(\log u)=\frac{d}{d x}[x \cdot \log (\sin x)]$
$\left.\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \log (\sin \mathrm{x})\right)+\log (\sin \mathrm{x}) \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\sin \mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})+\log (\sin \mathrm{x}) \cdot(1)\right]$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=(\sin \mathrm{x})^{\mathrm{x}}\left[\frac{\mathrm{x}}{\sin \mathrm{x}} \cdot \cos \mathrm{x}+\log (\sin \mathrm{x}) \cdot(1)\right]$
$\Rightarrow \frac{d u}{d x}=(\sin x)^{x}[x \cdot \cot x+\log \sin x]$
For $, v = \sin ^{-1} \sqrt{x}$
Now, differentiating both sides with respect $x$
$\frac{d v}{d x}=\frac{d}{d x}\left[\sin ^{-1} \sqrt{x}\right]$
$\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x})$
$\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2(\sqrt{x})}$
$\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^{2}}}$
Because $, y = u + v$
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\Rightarrow \frac{d y}{d x}=(\sin x)^{x}[x \cdot \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}$

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