Question

$\frac{d}{dx}\left[{\sin }^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)\right]$ w.r.t to $x$ equals

• A. $\frac{1}{2\sqrt{1-x^2}}$
• B. $\frac{-2}{\sqrt{1-x^2}}$
• C. $\frac{-1}{2\sqrt{1-x^2}}$
• D. None of these

Answer 1

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

Let $y={\sin }^{-1}(\frac{\sqrt{1+x}}{2}+\frac{\sqrt{1-x}}{2})$

by putting $x=\cos \theta$

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

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

$y={\sin }^{-1}\left(\sin (\frac{\pi }{4}+\frac{\theta }{2})\right)$

$y=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}x,y^{\prime }=\frac{1}{2\sqrt{1-x^2}}$