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