Question

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

Answer 1

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

Differentiating it with respect to x,

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

Differentiating it with respect to x,

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