Question

Differentiate

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

Answer 1

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

Differentiate it with respect to x,

$$\begin{gathered} \frac{dy}{dx}=0+1 \\ \therefore \frac{dy}{dx}=1 \end{gathered}$$