Question

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

Answer 1

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

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=\frac{2}{1+x^2}...\left(\mathrm{iii}\right) \\ \mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right), \\ v=2\theta \left[\mathrm{Since},{\cos }^{-1}\left(\cos \theta \right)=\theta ,\mathrm{if}\theta \in \left[0,\mathrm{\pi }\right]\right] \\ \Rightarrow v=2{\tan }^{-1}x\left[\mathrm{Since},x=\tan \theta \right] \end{gathered}$$

Differentiating it with respect to x,

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