Question

If $y={\sin }^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$, then the value of $\frac{dy}{dx}$ at $x=0$ is

• A. $1/2$
• B. $-1/2$
• C. $0$
• D. None of these

Answer 1

Answer: (B)

$-1/2$

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

Put $x=\cos 2\theta$

$\therefore y={\sin }^{-1}\left(\frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{2}\right)$

$={\sin }^{-1}\left(\frac{\sqrt{2}\cos \theta +\sqrt{2}\sin \theta }{2}\right)$

$={\sin }^{-1}\left(\sin (\frac{\pi }{4}+\theta )\right)$

$\Rightarrow y=\frac{\pi }{4}+\theta =\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}x$

$\Rightarrow \frac{dy}{dx}=0-\frac{1}{2\sqrt{1-x^2}}$

At $x=0$

$\frac{dy}{dx}=-\frac{1}{2}$