Question

Differentiate ${\tan }^{-1}\left(\frac{1-x}{1+x}\right)$ with respect to $\sqrt{1-x^2},\mathrm{if}-1<x<1$

Answer 1

$$\begin{gathered} \mathrm{Let},u={\tan }^{-1}\left(\frac{1-x}{1+x}\right) \\ \mathrm{Put}x=\tan \theta \\ \Rightarrow \theta ={\tan }^{-1}x \\ \Rightarrow u={\tan }^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right) \\ \Rightarrow u={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}-\theta \right)\right]...\left(\mathrm{i}\right) \\ \mathrm{Here}, \\ -1<x<1 \\ \Rightarrow -1<\tan \theta <1 \\ \Rightarrow -\frac{\mathrm{\pi }}{4}<\theta <\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{\mathrm{\pi }}{4}>-\theta >\frac{\mathrm{\pi }}{4} \\ \Rightarrow -\frac{\mathrm{\pi }}{4}<-\theta <\frac{\mathrm{\pi }}{4} \\ \Rightarrow 0<\frac{\mathrm{\pi }}{4}-\theta <\frac{\mathrm{\pi }}{2} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ u=\frac{\mathrm{\pi }}{4}-\theta \left[\mathrm{Since},{\tan }^{-1}\left(\tan \theta \right)=\theta ,\mathrm{if}\theta \in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\right] \\ \Rightarrow u=\frac{\mathrm{\pi }}{4}-{\tan }^{-1}x \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=0-\left(\frac{1}{1+x^2}\right) \\ \Rightarrow \frac{du}{dx}=-\frac{1}{1+x^2}...\left(\mathrm{ii}\right) \\ \mathrm{And}\mathrm{let},v=\sqrt{1-x^2} \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{dv}{dx}=\frac{1}{2\sqrt{1-x^2}}\times \frac{d}{dx}\left(1-x^2\right) \\ \Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{1-x^2}}\left(-2x\right) \\ \Rightarrow \frac{dv}{dx}=\frac{-x}{\sqrt{1-x^2}}...\left(\mathrm{iii}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}\left(\mathrm{iii}\right), \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=-\frac{1}{1+x^2}\times \frac{\sqrt{1-x^2}}{-x} \\ \therefore \frac{du}{dv}=\frac{\sqrt{1-x^2}}{x\left(1+x^2\right)} \end{gathered}$$