The problem asks us to find the derivative of the function $y = \sqrt{\sin^{-1}(x)}$.

1. Problem Description

The problem asks us to find the derivative of the function $y = \sqrt{\sin^{-1}(x)}$.

2. Solution Steps

We need to find $\frac{dy}{dx}$ where $y = \sqrt{\sin^{-1}(x)}$. We can rewrite $y$ as $y = (\sin^{-1}(x))^{\frac{1}{2}}$.

We will use the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$,

where $u = \sin^{-1}(x)$.

First, let's find $\frac{dy}{du}$. We have $y = u^{\frac{1}{2}}$.

$\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}}$.

Next, let's find $\frac{du}{dx}$. We have $u = \sin^{-1}(x)$.

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

Now, we can find $\frac{dy}{dx}$ using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{1}{\sqrt{1-x^2}}$.

Substitute $u = \sin^{-1}(x)$ back into the equation:

$\frac{dy}{dx} = \frac{1}{2\sqrt{\sin^{-1}(x)}} \cdot \frac{1}{\sqrt{1-x^2}}$.

Therefore,

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

3. Final Answer

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