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

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: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}\sqrt{1-x}}. \]