Question

If $y={\sin }^{-1}\left(\frac{x}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right),0<x<\infty$, prove that $\frac{dy}{dx}=\frac{2}{1+x^2}.$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)+{\cos }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \\ \mathrm{Put}x=\tan \theta \\ \therefore y={\sin }^{-1}\left(\frac{\tan \theta }{\sqrt{1+{\tan }^2\theta }}\right)+{\cos }^{-1}\left(\frac{1}{\sqrt{1+{\tan }^2\theta }}\right) \\ \Rightarrow y={\sin }^{-1}\left(\frac{{\displaystyle \frac{\sin \theta }{\cos \theta }}}{sec\theta }\right)+{\cos }^{-1}\left(\frac{1}{sec\theta }\right) \\ \Rightarrow y={\sin }^{-1}\left(\frac{{\displaystyle \frac{\sin \theta }{\cos \theta }}}{{\displaystyle \frac{1}{\cos \theta }}}\right)+{\cos }^{-1}\left(\cos \theta \right) \\ \Rightarrow y={\sin }^{-1}\left(\sin \theta \right)+{\cos }^{-1}\left(\cos \theta \right)...\left(\mathrm{i}\right) \\ \mathrm{Here},0<x<\infty \\ \Rightarrow 0<\tan \theta <\infty \\ \Rightarrow 0<\theta <\frac{\mathrm{\pi }}{2} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ y=\theta +\theta \left[\begin{array}{c}\mathrm{Since},{\sin }^{-1}\left(\sin \theta \right)=\theta ,\mathrm{if}\theta \in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\\ {\cos }^{-1}\left(\cos \theta \right)=\theta ,\mathrm{if}\theta \in \left[0,\mathrm{\pi }\right]\end{array}\right] \\ \Rightarrow y=2\theta \\ \Rightarrow y=2{\tan }^{-1}x\left[\mathrm{Since},x=\tan \theta \right] \end{gathered}$$

Differentiate it with respect to x,

$\therefore \frac{dy}{dx}=\frac{2}{1+x^2}$