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

Answer 1

The correct option is C

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

Finding the value of $\frac{dy}{dx}$:

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

Put $x=\sin \theta$

$\Rightarrow \theta ={\sin }^{-1}x$

$$\begin{gathered} y={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right) \\ ={\sin }^{-1}\left(2\sin \theta \sqrt{{\cos }^2\theta }\right)\left[\because {\cos }^2\theta +{\sin }^2\theta =1\right] \\ ={\sin }^{-1}\left(2\sin \theta \cos \theta \right) \\ ={\sin }^{-1}\left(\sin 2\theta \right) \\ =2\theta \\ =2{\sin }^{-1}x \end{gathered}$$

So, $y=2{\sin }^{-1}x$

Differentiating with respect to $x$ both sides:

$\begin{array}{rcl}\frac{dy}{dx}& =& 2\times \frac{1}{\sqrt{1-x^2}}\\ & =& \frac{2}{\sqrt{1-x^2}}\end{array}$

Hence, option (C) is the correct answer.