Differentiate the following functions with respect to x :

Let $y=\sin \left(x^{x}\right)$

Take sin inverse both sides:

$\Rightarrow \sin ^{-1} y=\sin ^{-1}\left(\sin x^{x}\right)$

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

Taking log both the sides:

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

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

Differentiating with respect to $\mathrm{x}$ :

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

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

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

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

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

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

$\left\{\frac{d\left(\sin ^{-1} u\right)}{d x}=\frac{1}{\sqrt{1-u^{2}}} \frac{d u}{d x}\right\}$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin ^{-1} \mathrm{y}\left(\sqrt{1-\mathrm{y}^{2}}\right)(1+\log \mathrm{x})$

Put the value of $y=\sin \left(x^{x}\right):$

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

$\Rightarrow \frac{d y}{d x}=x^{x}\left(\sqrt{\cos ^{2}\left(x^{x}\right)}\right)(1+\log x)$

$\left\{\sin ^{2} x+\cos ^{2} x=1\right\}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{\mathrm{x}} \cos \mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})$