Question

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

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

Answer 1

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

Put $ax=\sin \theta ,\theta \in (-\frac{\pi }{4},\frac{\pi }{4})$
$y={\sin }^{-1}\left[2\sin \theta \sqrt{1-{\sin }^2\theta }\right]$
$={\sin }^{-1}\left[2\sin \theta \cos \theta \right]$
$={\sin }^{-1}\left[\sin 2\theta \right]$
$(\because 2\theta \in (-\frac{\pi }{2},\frac{\pi }{2}))$
$y=2\theta =2{\sin }^{-1}ax$
$\Rightarrow$$\frac{dy}{dx}=2\times \frac{a}{\sqrt{1-a^2{x}^2}}$
$\Rightarrow$$\frac{dy}{dx}=\frac{2a}{\sqrt{1-a^2{x}^2}}$