Question

Differentiate

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

Answer 1

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

Differentiate it with respect to x,

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