Question

Differentiate ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ with respect to $x$, if
$-1<x<-\frac{1}{\sqrt{2}}$

Answer 1

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

On differentiating both sides w.r.t $x$, we have

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

We know that,

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

Therefore,

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

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

$\frac{dy}{dx}=\frac{1}{\sqrt{1-\left(4x^2(1-x^2)\right)}}\times (2\sqrt{1-x^2}+\frac{2x}{2\sqrt{1-x^2}}\times (0-2x))$

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

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

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

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

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

Hence, this is the answer.