Find the derivative of the function $$ y=\sin ^{-1}(\sqrt{x}) $$

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1. We start with the function:
   $$
   y = \sin^{-1}(\sqrt{x})
   $$

2. To find the derivative $$ \frac{dy}{dx} $$, we use the chain rule. Recall that if
   $$
   y = \sin^{-1}(u),
   $$
   then
   $$
   \frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}}.
   $$

3. In our problem, let:
   $$
   u = \sqrt{x}.
   $$
   Then the derivative of $$ y $$ with respect to $$ u $$ is:
   $$
   \frac{dy}{du} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} = \frac{1}{\sqrt{1 - x}}.
   $$

4. Next, compute the derivative of $$ u = \sqrt{x} $$ with respect to $$ x $$:
   $$
   u = x^{\frac{1}{2}} \quad \Rightarrow \quad \frac{du}{dx} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}.
   $$

5. Now, applying the chain rule:
   $$
   \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}}.
   $$

6. Therefore, the derivative of the function is:
   $$
   \frac{dy}{dx} = \frac{1}{2\sqrt{x}\sqrt{1-x}}.
   $$
The derivative of $$ y = \sin^{-1}(\sqrt{x}) $$ is $$ \frac{dy}{dx} = \frac{1}{2\sqrt{x}\sqrt{1-x}} $$.

Find the derivative of the function \( y=\sin ^{-1}(\sqrt{x}) \)

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