Question

Differentiate

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

Answer 1

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

Differentiate it with respect to x,

$$\begin{gathered} \frac{dy}{dx}=0+\frac{1}{1+x^2} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2} \end{gathered}$$