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

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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