Question

Differentiate

${\tan }^{-1}\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}}\right)$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\tan }^{-1}\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}}\right) \\ \Rightarrow y={\tan }^{-1}\sqrt{x}+{\tan }^{-1}\sqrt{a}\left[\mathrm{Since},{\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}\right] \end{gathered}$$

Differentiate it with respect to x using chain rule,

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left({\tan }^{-1}\sqrt{x}\right)+\frac{d}{dx}\left({\tan }^{-1}\sqrt{a}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+{\left(\sqrt{x}\right)}^2}\frac{d}{dx}\left(\sqrt{x}\right)+0 \\ \Rightarrow \frac{dy}{dx}=\left(\frac{1}{1+x}\right)\left(\frac{1}{2\sqrt{x}}\right) \\ \therefore \frac{dy}{dx}=\frac{1}{2\sqrt{x}\left(1+x\right)} \end{gathered}$$