Question

Differentiate ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ with respect to ${\sec }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)$, if
(i) $x\in \left(0,\frac{1}{\sqrt{2}}\right)$

(ii) $x\in \left(\frac{1}{\sqrt{2}},1\right)$

Answer 1

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

Differentiating it with respect to x,

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

Differentiating it with respect to x,

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


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

Differentiating it with respect to x,

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

Differentiating it with respect to x,

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