Question

The derivative of ${\sin }^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ with respect to $x$ is

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

Answer 1

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

Let $y={\sin }^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$
On putting $x=\cos \theta$, we get
$y={\sin }^{-1}(\frac{1}{\sqrt{2}}\cos \frac{\theta }{2}+\frac{1}{\sqrt{2}}\sin \frac{\theta }{2})$
$\Rightarrow y={\sin }^{-1}\left\{\sin (\frac{\pi }{4}+\frac{\theta }{2})\right\}$
$\Rightarrow y=\frac{\pi }{4}+\frac{\theta }{2}$
$\Rightarrow y=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}x$
$\therefore \frac{dy}{dx}=-\frac{1}{2\sqrt{1-x^2}}$