MCQ
If $y =\sin ^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right]$, then $\frac{ dy }{ dx }=$ ?
- ✓$\left(-\frac{1}{2}\right) \frac{1}{\sqrt{1-x^2}}$
- B$\left(-\frac{1}{2}\right) \frac{1}{\sqrt{x^2-1}}$
- C$\left(\frac{1}{4}\right) \frac{1}{\sqrt{x^2-1}}$
- D$\left(\frac{1}{4}\right) \frac{1}{\sqrt{1-x^2}}$