If $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2},x∈ (-1,0)$, then $\frac{dy}{dx}$ is equal to

$\frac{2}{\sqrt{1-x^2}}$

The correct answer is Option (3) → $\frac{2}{\sqrt{1-x^2}}$

Given $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2},\;x\in(-1,0)$.

$\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$.

For $u=\sqrt{1-x^2}$, $u'=\frac{d}{dx}\sqrt{1-x^2}=\frac{-x}{\sqrt{1-x^2}}$ and $\sqrt{1-u^2}=\sqrt{1-(1-x^2)}=\sqrt{x^2}=|x|=-x$ (since $x\in(-1,0)$).

Hence $\frac{d}{dx}\sin^{-1}\sqrt{1-x^2}=\frac{u'}{\sqrt{1-u^2}} =\frac{-x/\sqrt{1-x^2}}{-x}=\frac{1}{\sqrt{1-x^2}}$.

Therefore,

$\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$