Question

Differentiate

$\sqrt{{\tan }^{-1}\left(\frac{x}{2}\right)}$

Answer 1

$$\begin{gathered} \mathrm{Let}y=\sqrt{{\tan }^{-1}\left(\frac{x}{2}\right)} \\ \Rightarrow y={\left\{{\tan }^{-1}\left(\frac{x}{2}\right)\right\}}^{\frac{1}{2}} \end{gathered}$$
Differentiate it with respect to x we get,
$$\begin{gathered} \frac{\mathit{d}y}{\mathit{d}x}=\frac{d}{dx}{\left\{{\tan }^{-1}\left(\frac{x}{2}\right)\right\}}^{\frac{1}{2}} \\ =\frac{1}{2}{\left\{{\tan }^{-1}\left(\frac{x}{2}\right)\right\}}^{\frac{1}{2}-1}\frac{d}{dx}\left({\tan }^{-1}\frac{x}{2}\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right] \\ =\frac{1}{2}{\left\{{\tan }^{-1}\left(\frac{x}{2}\right)\right\}}^{\frac{-1}{2}}\times \frac{1}{1+{\left({\displaystyle \frac{x}{2}}\right)}^2}\times \frac{d}{dx}\left(\frac{x}{2}\right) \\ =\frac{4}{4\left(4+x^2\right)\sqrt{{\tan }^{-1}\left({\displaystyle \frac{x}{2}}\right)}} \\ =\frac{1}{\left(4+x^2\right)\sqrt{{\tan }^{-1}\left({\displaystyle \frac{x}{2}}\right)}} \\ \mathrm{So},\frac{d}{dx}\left\{\sqrt{{\tan }^{-1}\left(\frac{x}{2}\right)}\right\}=\frac{1}{\left(4+x^2\right)\sqrt{{\tan }^{-1}\left({\displaystyle \frac{x}{2}}\right)}} \end{gathered}$$