Question

Differentiate ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$, if $-1<x<1,x\ne 0.$

Answer 1

$$\begin{gathered} \mathrm{Let},u={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \\ \mathrm{put}x=\tan \theta \\ \Rightarrow u={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right) \\ \Rightarrow u={\tan }^{-1}\left(\frac{sec\theta -1}{\tan \theta }\right) \\ \Rightarrow u={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right) \\ \Rightarrow u={\tan }^{-1}\left(\frac{2{\sin }^2{\displaystyle \frac{\theta }{2}}}{2{\displaystyle \sin \frac{\theta }{2}}{\displaystyle \cos \frac{{\displaystyle \theta }}{2}}}\right) \\ \Rightarrow u={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)...\left(\mathrm{i}\right) \\ \mathrm{And}, \\ v={\sin }^{-1}\left(\frac{2x}{1+x^2}\right) \\ \Rightarrow v={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right) \\ \Rightarrow v={\sin }^{-1}\left(\sin 2\theta \right)...\left(\mathrm{ii}\right) \\ \mathrm{Here}, \\ -1<x<1 \\ \Rightarrow -1<\tan \theta <1 \\ \Rightarrow -\frac{\mathrm{\pi }}{4}<\theta <\frac{\mathrm{\pi }}{4}...\left(\mathrm{A}\right) \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ u=\frac{\theta }{2}\left[\mathrm{Since},{\tan }^{-1}\left(\tan \theta \right)=\theta ,\mathrm{if}\theta \in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\right] \\ \Rightarrow u=\frac{1}{2}{\tan }^{-1}x\left[\mathrm{since},x=\tan \theta \right] \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=\frac{1}{2}\left(\frac{1}{1+x^2}\right) \\ \Rightarrow \frac{du}{dx}=\frac{1}{2\left(1+x^2\right)}...\left(\mathrm{i}\right) \\ \mathrm{Now},\mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{A}\right), \\ v=2\theta \left[\mathrm{Since},{\sin }^{-1}\left(\sin \theta \right)=\theta ,\mathrm{if}\theta \in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\right] \\ \Rightarrow v=2{\tan }^{-1}x\left[\mathrm{Since},x=\tan \theta \right] \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{dv}{dx}=2\left(\frac{1}{1+x^2}\right)...\left(\mathrm{iv}\right) \\ \mathrm{dividing}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{by}\left(\mathrm{iv}\right), \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{1}{2\left(1+x^2\right)}\times \frac{1+x^2}{2} \\ \therefore \frac{du}{dv}=\frac{1}{4} \end{gathered}$$