Question

$\frac{d}{dx}\left({\sin }^{-1}[x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2}]\right)$

• A. $\frac{1}{\sqrt{(1-x^2)}}+\frac{1}{\sqrt{(1-x)}}\cdot \frac{1}{2\sqrt{\left(x\right)}}\cdot$
• B. $\frac{1}{\sqrt{(1+x^2)}}+\frac{1}{\sqrt{(1-x)}}\cdot \frac{1}{2\sqrt{\left(x\right)}}\cdot$
• C. $\frac{1}{\sqrt{(1-x^2)}}+\frac{1}{\sqrt{(1+x)}}\cdot \frac{1}{2\sqrt{\left(x\right)}}\cdot$
• D. $\frac{1}{\sqrt{(1+x^2)}}+\frac{1}{\sqrt{(1+x)}}\cdot \frac{1}{2\sqrt{\left(x\right)}}\cdot$

Answer 1

The correct option is A $\frac{1}{\sqrt{(1-x^2)}}+\frac{1}{\sqrt{(1-x)}}\cdot \frac{1}{2\sqrt{\left(x\right)}}\cdot$

Let $y={\sin }^{-1}[x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2}]$

Put $x=\sin \theta ,$ and $\sqrt{x}=\sin \varphi$

$\therefore y={\sin }^{-1}(\sin \theta \cos \varphi +\cos \theta \sin \varphi )$

$={\sin }^{-1}\sin (\theta +\varphi )=\theta +\varphi$

or $y={\sin }^{-1}x+{\sin }^{-1}\sqrt{x}$

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