What is $\frac{dy}{dx}$ for $y = \sin^{-1}\left(2x\sqrt{1 - x^{2}}\right)$, $\lvert x \rvert < \frac{1}{\sqrt{2}}$?

Explanation

1. 01

   Put $x = \sin \theta$: $2\sin \theta\cos \theta = \sin 2\theta$, so the function is $2\theta = 2\sin^{-1} x$.

2. 02

   So $\frac{dy}{dx} = \frac{2}{\sqrt{1 - x^{2}}}$.