Find $\large \frac {dy}{dx} $ in the following: $y = \sin^{-1} \big( 2x \sqrt{1 -x^2} \big), -\;{ \frac{1}{\sqrt2} }\;< x < \;{\frac{1}{\sqrt2} }\\ $

Question

$\begin{array}{1 1} \large\frac{dy}{dx}=\large\frac{-2}{\sqrt{1-x^2}} \\ \large\frac{dy}{dx}=\large\frac{2}{\sqrt{1+x^2}} \\ \large\frac{dy}{dx}=\large\frac{2}{\sqrt{1-x^3}} \\ \large\frac{dy}{dx}=\large\frac{2}{\sqrt{1-x^2}} \end{array} $