Differentiate the following functions with respect to x :

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

Taking log both the sides:

$\Rightarrow \log y=\log (\sin x)^{\log x}$

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

Differentiating with respect to $x$ :

$\Rightarrow \frac{d(\log y)}{d x}=\frac{d(\log x \log \sin x)}{d x}$

$\Rightarrow \frac{d(\log y)}{d x}=\log x \times \frac{d(\log \sin x)}{d x}+\log \sin x \times \frac{d(\log x)}{d x}$

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

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

$\left\{\frac{d(\log u)}{d x}=\frac{1}{u} \frac{d u}{d x} ; \frac{d(\sin x)}{d x}=\cos x\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{\log x}{\sin x}(\cos x)+\frac{\log \sin x}{x}$

$\Rightarrow \frac{d y}{d x}=y\left\{\log x \cot x+\frac{\log \sin x}{x}\right\}$

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

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