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

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

$\Rightarrow \sin y=2x\sqrt{1-x^2}$

Differentiating this relationship w.r.t $x$ we get

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

$\Rightarrow \sqrt{1-{\sin }^{2}y}\frac{dy}{dx}=2[\frac{x}{2}\frac{-2x}{\sqrt{1-x^2}}+\sqrt{1-x^2}]$

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

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

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

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