Question

Differentiate:$y={sin}^{-1}\left(2x\sqrt{1-x^2}\right)$

Answer 1

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

$\frac{dy}{dx}=\frac{1}{\sqrt{1-4x^2(1-x^2)}}\frac{d}{dx}\left(2x\sqrt{1-x^2}\right)$

$=\frac{2}{\sqrt{1-4x^2+4x^4}}[x\frac{d}{dx}\left(\sqrt{1-x^2}\right)+\sqrt{1-x^2}\frac{d}{dx}\left(x\right)]$

$=\frac{2}{\sqrt{{\left(2x^2\right)}^2-2\times 2x^2+1}}[\frac{-2x^2}{2\sqrt{1-x^2}}+\sqrt{1-x^2}]$

$=\frac{2}{\sqrt{{(2x^2-1)}^2}}\left[\frac{-x^2+1-x^2}{\sqrt{1-x^2}}\right]$

$=\frac{2}{2x^2-1}\times \frac{-(2x^2-1)}{\sqrt{1-x^2}}$

$=\frac{-2}{\sqrt{1-x^2}}$