Question

Differentiate ${\sin }^{-1}\frac{\sqrt{1+x}+\sqrt{1-x}}{2}.$

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

Answer 1

Answer: (C)

$\frac{-1}{2\sqrt{(1-x^2)}}$

Let $y={\sin }^{-1}\frac{\sqrt{1+x}+\sqrt{1-x}}{2}.$
Put $x=\cos \theta \Rightarrow 1+\cos \theta =2{\cos }^2\left(\theta /2\right),$
and $1-\cos \theta =2{\sin }^2\left(\theta /2\right)$
$\therefore y={\sin }^{-1}\frac{\cos \left(\theta /2\right)+\sin \left(\theta /2\right)}{\sqrt{2}}$
$={\sin }^{-1}\sin (\theta /2+\pi /4)=\theta /2+\pi /4$
$\therefore y=\frac{1}{2}{\cos }^{-1}x+\frac{\pi }{4}$
$\frac{dy}{dx}=\frac{-1}{2\sqrt{(1-x^2)}}$