Question

If $y={\sin }^{-1}\sqrt{1-x}+{\cos }^{-1}\sqrt{x}$, then $\frac{dy}{dx}$ equals

• A. $\frac{1}{\sqrt{x(1-x)}}$
• B. $\frac{-1}{\sqrt{x(1-x)}}$
• C. $\frac{1}{\sqrt{x(a+x)}}$
• D. none of these

Answer 1

Answer: (B)

$y={\sin }^{-1}\sqrt{1-x}+{\cos }^{-1}\sqrt{x}$
$\therefore \frac{dy}{dx}=\frac{1}{\sqrt{1-\left(1-x\right)}}.\frac{d}{dx}\sqrt{1-x}-\frac{1}{\sqrt{1-x}}\frac{d}{dx}\left(\sqrt{x}\right)$
$=\frac{1}{\sqrt{x}}.\frac{1\left(-1\right)}{2\sqrt{1-x}}-\frac{1}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}}$
$=-\frac{1}{2\sqrt{x}\sqrt{1-x}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}$
$=-\frac{1}{\sqrt{x}\sqrt{1-x}}=-\frac{1}{\sqrt{x\left(1-x\right)}}$