Question

If $y={\sin }^{-1}\left(2x\sqrt{1-x^2}\right),-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{x}{\sqrt{1-x^2}}$
• B. $\frac{1}{\sqrt{1-x^2}}$
• C. $\frac{2}{\sqrt{1-x^2}}$
• D. $\frac{2x}{\sqrt{1-x^2}}$
• E. $\frac{-2x}{\sqrt{1-x^2}}$

Answer 1

The correct option is C $\frac{2}{\sqrt{1-x^2}}$

Given, $y={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$
Put $x=\sin \theta$
$\therefore y={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)$
$={\sin }^{-1}\left(2\sin \theta \cos \theta \right)$
$={\sin }^{-1}\left(\sin 2\theta \right)=2\theta$
$\Rightarrow y=2{\sin }^{-1}x$
On differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$