Differentiate the following functions with respect to x :

Let $y=\log \left(\cos x^{2}\right)$

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

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

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

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

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

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

$\Rightarrow \frac{d y}{d x}=-\frac{\sin x^{2}}{\cos x^{2}} \frac{d}{d x}\left(x^{2}\right)$

$\Rightarrow \frac{d y}{d x}=-\tan x^{2} \frac{d}{d x}\left(x^{2}\right)$

However, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\tan \mathrm{x}^{2} \times 2 \mathrm{x}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=-2 \mathrm{x} \tan \mathrm{x}^{2}$

Thus, $\frac{d}{d x}\left[\log \left(\cos x^{2}\right)\right]=-2 x \tan x^{2}$