Question

Find $\frac{dy}{dx}$, if $y={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$, $\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$.

Answer 1

$\begin{array}{c}Put\\ x=\sin \theta \\ \theta ={\sin }^{-1}x\\ \frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}<\sin \theta <\frac{1}{\sqrt{2}}\\ {\sin }^{-1}\frac{-1}{\sqrt{2}}<\theta <{\sin }^{-1}\frac{1}{\sqrt{2}}\\ \frac{-\pi }{4}<\theta <\frac{\pi }{4}\\ \frac{-\pi }{2}<2\theta <\frac{\pi }{2}\\ 2\theta \in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)\\ \therefore {\sin }^{-1}\left(\sin 2\theta \right)=2\theta \\ {\sin }^{-1}\left(2\sin \theta \cos \theta \right)=2\theta \\ {\sin }^{-1}\left(2\times x\sqrt{1-x^2}\right)=2{\sin }^{-1}x\\ y=2{\sin }^{-1}x\\ \frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}\end{array}$