Question

Differentiate

${\cos }^{-1}\left\{2x\sqrt{1-x^2}\right\},\frac{1}{\sqrt{2}}<x<1$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\cos }^{-1}\left\{2x\sqrt{1-x^2}\right\} \\ \mathrm{Put}x=\cos \theta \\ y={\cos }^{-1}\left\{2\cos \theta \sqrt{1-{\cos }^2\theta }\right\} \\ y={\cos }^{-1}\left\{2\cos \theta \sin \theta \right\} \\ y={\cos }^{-1}\left\{\sin 2\theta \right\}\left[\mathrm{Since},\sin 2\theta =2\sin \theta \cos \theta \right] \\ y={\cos }^{-1}\left[\cos \left(\frac{\mathrm{\pi }}{2}-2\theta \right)\right]...\left(\mathrm{i}\right) \\ \mathrm{Now}, \\ \frac{1}{\sqrt{2}}<x<1 \\ \Rightarrow \frac{1}{\sqrt{2}}<\cos \theta <1 \\ \Rightarrow 0<\theta <\frac{\mathrm{\pi }}{4} \\ \Rightarrow 0<2\theta <\frac{\mathrm{\pi }}{2} \\ \Rightarrow 0>-2\theta >-\frac{\mathrm{\pi }}{2} \\ \Rightarrow \frac{\mathrm{\pi }}{2}>\left(\frac{\mathrm{\pi }}{2}-2\theta \right)>0 \\ \mathrm{Hence},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right) \\ y=\frac{\mathrm{\pi }}{2}-2\theta \left[\mathrm{Since},{\cos }^{-1}\left(\cos \theta \right)=\theta ,\mathrm{if}\theta \in \left[0,\mathrm{\pi }\right]\right] \\ y=\frac{\mathrm{\pi }}{2}-2{\cos }^{-1}x\left[\mathrm{Since},x=\cos \theta \right] \\ \mathrm{differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\mathrm{\pi }}{2}\right)-2\frac{d}{dx}\left({\cos }^{-1}x\right) \\ \frac{dy}{dx}=0-2\left(\frac{-1}{\sqrt{1-x^2}}\right) \\ \frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}} \end{gathered}$$