Q. $\frac{d}{dx}\sin^{- 1}{\ (x\sqrt{1 - x} + \sqrt{x}\sqrt{1 - x^{2})}\ }$is equal to
A) $- \frac{1}{2x\sqrt{1 - x}} - \frac{1}{\sqrt{1 - x^{2}}}$
B) $\frac{1}{2\sqrt{x}\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - x^{2}}}$
C) $\frac{1}{2\sqrt{x}\sqrt{1 - x}} + \frac{1}{\sqrt{1 - x^{2}}}$
D) $- \frac{1}{2\sqrt{\ x}\sqrt{1 - x}} + \frac{1}{\sqrt{1 - x^{2}}}$
Explanation
(c)
Since, $\sin^{- 1}{\left( x\sqrt{1 - x} + \sqrt{x}\sqrt{1 - x^{2}} \right) = \sin^{- 1}{x + \sin^{- 1}\sqrt{x}}}$
$\therefore\ \ \ \ \ \frac{d}{dx}\{\sin^{- 1}{(x\sqrt{1 - x} + \sqrt{x}\sqrt{1 - x^{2}})\}}$
$= \frac{d}{dx}(\sin^{- 1}{x + \sin^{- 1}{\sqrt{x}) = \frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{2\sqrt{x}\sqrt{1 - x}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.