Question

Derivative of ${\sin }^{-1}\sqrt{1-x}+{\cos }^{-1}\sqrt{x}.$ is

• A. $-\frac{1}{\sqrt{\left(x\right)}\sqrt{(1-x)}}.$
• B. $-\frac{1}{\sqrt{\left(x\right)}\sqrt{(1+x)}}.$
• C. $\frac{1}{\sqrt{\left(x\right)}\sqrt{(1-x)}}.$
• D. $\frac{1}{\sqrt{\left(x\right)}\sqrt{(1+x)}}.$

Answer 1

The correct option is A $-\frac{1}{\sqrt{\left(x\right)}\sqrt{(1-x)}}.$

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