Differentiate the following functions with respect to x :

Solution:

Let $y=\sqrt{\tan ^{-1} \frac{x}{2}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{\tan ^{-1} \frac{\mathrm{x}}{2}}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)^{\frac{1}{2}}\right]$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)^{\frac{1}{2}-1} \frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)$ [using chain rule]

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)^{-\frac{1}{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\tan ^{-1} \frac{x}{2}\right)^{-\frac{1}{2}} \frac{1}{1+\left(\frac{x}{2}\right)^{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{x}{2}\right)$ [using chain rule]

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)^{-\frac{1}{2}} \frac{1}{1+\frac{\mathrm{x}^{2}}{4}} \times \frac{1}{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)^{-\frac{1}{2}} \frac{4}{4+\mathrm{x}^{2}} \times \frac{1}{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)^{-\frac{1}{2}} \frac{1}{4+\mathrm{x}^{2}} \times \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$

However, $\frac{d}{d x}(x)=1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)^{-\frac{1}{2}} \frac{1}{4+\mathrm{x}^{2}} \times 1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\tan ^{-1} \frac{\mathrm{x}}{2}\right)^{-\frac{1}{2}} \frac{1}{4+\mathrm{x}^{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\left(4+x^{2}\right)\left(\tan ^{-1} \frac{x}{2}\right)^{\frac{1}{2}}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\left(4+\mathrm{x}^{2}\right) \sqrt{\tan ^{-1} \frac{\mathrm{x}}{2}}}$

Thus, $\frac{d}{d x}\left(\sqrt{\tan ^{-1} \frac{x}{2}}\right)=\frac{1}{\left(4+x^{2}\right) \sqrt{\tan ^{-1} \frac{x}{2}}}$