Question

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

• A. $\frac{1}{\sin \sqrt{a-ax}}$
• B. $\sin \sqrt{x}.\sin \sqrt{a}$
• C. $\frac{1}{2\sqrt{x}\sqrt{1-x}}$
• D. $0$

Answer 1

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

Given $y={\sin }^{-1}[\sqrt{x-ax}-\sqrt{a-ax}]$

Put $x={\sin }^2\theta ,a={\sin }^2\varphi$ then

$y={\sin }^{-1}[\sqrt{{\sin }^2\theta -{\sin }^2\varphi {\sin }^2\theta }-\sqrt{{\sin }^2\varphi -{\sin }^2\varphi {\sin }^2\theta }]$

$={\sin }^{-1}[\sqrt{{\sin }^2\theta (1-{\sin }^2\varphi )}-\sqrt{{\sin }^2\varphi (1-{\sin }^2\theta )}]$

$={\sin }^{-1}[\sin \theta \cos \varphi -\sin \varphi \cos \theta ]$

$={\sin }^{-1}\left[\sin (\theta -\varphi )\right]=\theta -\varphi$

$={\sin }^{-1}\sqrt{x}-{\sin }^{-1}\sqrt{a}$

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