Question

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

• 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

The Correct Option is B

The correct option is B) : $\frac{-1}{2\sqrt{1-x^2}}$.