Differentiate the following functions with respect to x :

Let $y=x^{\sin ^{-1} x}$

Taking log both the sides:

$\Rightarrow \log y=\log x^{\sin ^{-1} x}$

$\Rightarrow \log y=\sin ^{-1} x \log x\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $x$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\sin ^{-1} \mathrm{x} \log \mathrm{x}\right)}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\sin ^{-1} \mathrm{x} \times \frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}+\log \times \frac{\mathrm{d}\left(\sin ^{-1} \mathrm{x}\right)}{\mathrm{dx}}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\sin ^{-1} x \times \frac{1}{x} \frac{d x}{d x}+\log x \times \frac{1}{\sqrt{1-x^{2}}} \frac{d x}{d x}$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}} ; \frac{\mathrm{d}\left(\sin ^{-1} \mathrm{u}\right)}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{u}^{2}}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{\sin ^{-1} x}{x}+\frac{\log x}{\sqrt{1-x^{2}}}$

$\Rightarrow \frac{d y}{d x}=y\left\{\frac{\sin ^{-1} x}{x}+\frac{\log x}{\sqrt{1-x^{2}}}\right\}$

Put the value of $y=x^{\sin ^{-1} x}:$

$\Rightarrow \frac{d y}{d x}=x^{\sin ^{-1} x}\left\{\frac{\sin ^{-1} x}{x}+\frac{\log x}{\sqrt{1-x^{2}}}\right\}$