Question

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

Answer 1

Put $x=\sin \theta \Rightarrow \theta ={\sin }^{-1}x$

Then we have,

$y={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)={\sin }^{-1}\left(2\sin \theta \cos \theta \right)$

$={\sin }^{-1}\left(\sin 2\theta \right)=2\theta =2{\sin }^{-1}x$

$\therefore \frac{dy}{dx}=2\cdot \frac{1}{\sqrt{1-x^2}}=\frac{2}{\sqrt{1-x^2}}$