For $x \in(0,1)$, the derivative of $\sin ^{-1} x$ with respect to $\cos ^{-1} \sqrt{1-x^2}$ is:

1

The correct answer is Option (1) → 1

$y=\sin^{-1}x$, $z=\cos^{-1}\sqrt{1-x^2}$

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

$\frac{dz}{dx}=\frac{-1}{\sqrt{1-(1-x^2)}}×\frac{(-2x)}{2\sqrt{1-x^2}}$

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

so $\frac{dy}{dx}=\frac{d(\sin^{-1}x)}{d(\cos^{-1}\sqrt{1-x^2})}=1$