Differentiate the following functions with respect to x :

Let $y=(\log \sin x)^{2}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[(\log (\sin \mathrm{x}))^{2}\right]$

We know $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$

$\Rightarrow \frac{d y}{d x}=2(\log (\sin x))^{2-1} \frac{d}{d x}[\log (\sin x)]$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=2 \log (\sin x) \frac{d}{d x}[\log (\sin x)]$

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

$\Rightarrow \frac{d y}{d x}=2 \log (\sin x)\left[\frac{1}{\sin x} \frac{d}{d x}(\sin x)\right]$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=\frac{2}{\sin x} \log (\sin x) \frac{d}{d x}(\sin x)$

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

$\Rightarrow \frac{d y}{d x}=\frac{2}{\sin x} \log (\sin x) \cos x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2\left(\frac{\cos \mathrm{x}}{\sin \mathrm{x}}\right) \log (\sin \mathrm{x})$

$\therefore \frac{d y}{d x}=2 \cot x \log (\sin x)$

Thus, $\frac{d}{d x}\left[(\log (\sin x))^{2}\right]=2 \cot x \log (\sin x)$