Question

Differentiate

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

Answer 1

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

Differentiate it with respect to x,

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