Differentiate the following functions with respect to x :

Let $y=\sin (\log \sin x)$

On differentiating $y$ with respect to $x$, we get

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

We know $\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})=\cos \mathrm{x}$

$\Rightarrow \frac{d y}{d x}=\cos (\log (\sin x)) \frac{d}{d x}[\log (\sin x)]$ [using chain rule]

We have $\frac{d}{d x}(\log x)=\frac{1}{x}$

$\Rightarrow \frac{d y}{d x}=\cos (\log (\sin x))\left[\frac{1}{\sin x} \frac{d}{d x}(\sin x)\right]$ [using chain rule]

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

However, $\frac{d}{d x}(\sin x)=\cos x$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{\cos \mathrm{x}}{\sin \mathrm{x}}\right) \cos (\log (\sin \mathrm{x}))$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\cot \mathrm{x} \cos (\log (\sin \mathrm{x}))$

Thus, $\frac{d}{d x}[\sin (\log (\sin x))]=\cot x \cos (\log (\sin x))$