Differentiate the following functions with respect to x :

Let $y=\log \left(\tan ^{-1} x\right)$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\log \left(\tan ^{-1} \mathrm{x}\right)\right]$

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

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

We have $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\tan ^{-1} \mathrm{x}} \times \frac{1}{1+\mathrm{x}^{2}}$

$\therefore \frac{d y}{d x}=\frac{1}{\left(1+x^{2}\right) \tan ^{-1} x}$

Thus, $\frac{d}{d x}\left[\log \left(\tan ^{-1} x\right)\right]=\frac{1}{\left(1+x^{2}\right) \tan ^{-1} x}$