Question

Differentiate ${\cot }^{-1}\left(\frac{1-x}{1+x}\right)$ with respect to x.

Answer 1

$$\begin{gathered} \mathrm{Let},y=co{t}^{-1}\left(\frac{1-x}{1+x}\right) \\ \mathrm{Put}x=\tan \theta \\ \therefore y=co{t}^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right) \\ =co{t}^{-1}\left(\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\tan \theta }{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan \theta }\right) \\ =co{t}^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}-\theta \right)\right] \\ =co{t}^{-1}\left[cot\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}+\theta \right)\right] \\ =\frac{\mathrm{\pi }}{4}+\theta \\ =\frac{\mathrm{\pi }}{4}+{\tan }^{-1}x\left[\mathrm{Since},x=\tan \theta \right] \end{gathered}$$

Differentiate it with respect to x,

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