Find `(dy)/(dx)" if y"=sqrt(sin^(-1) sqrt(x))`.

Answer

To find \(\frac{dy}{dx}\) where \(y = \sqrt{\sin^{-1}(\sqrt{x})}\), we will use the chain rule and the derivatives of inverse trigonometric functions.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = \sqrt{\sin^{-1}(\sqrt{x})} \]

2. **Differentiate using the chain rule**:
The derivative of \(y\) with respect to \(x\) can be found using the chain rule. We first differentiate the outer function, which is the square root function:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{\sin^{-1}(\sqrt{x})}} \cdot \frac{d}{dx}[\sin^{-1}(\sqrt{x})] \]

3. **Differentiate the inner function**:
Now we need to differentiate \(\sin^{-1}(\sqrt{x})\). The derivative of \(\sin^{-1}(u)\) is given by \(\frac{1}{\sqrt{1 - u^2}}\), where \(u = \sqrt{x}\). Thus:
\[ \frac{d}{dx}[\sin^{-1}(\sqrt{x})] = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}[\sqrt{x}] \]

4. **Differentiate \(\sqrt{x}\)**:
The derivative of \(\sqrt{x}\) is:
\[ \frac{d}{dx}[\sqrt{x}] = \frac{1}{2\sqrt{x}} \]

5. **Combine the derivatives**:
Now substituting back, we have:
\[ \frac{d}{dx}[\sin^{-1}(\sqrt{x})] = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} \]

6. **Substitute back into the derivative of \(y\)**:
Now substituting this back into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{\sin^{-1}(\sqrt{x})}} \cdot \left(\frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}}\right) \]

7. **Simplify the expression**:
Combining everything, we get:
\[ \frac{dy}{dx} = \frac{1}{4\sqrt{x}\sqrt{\sin^{-1}(\sqrt{x})}\sqrt{1 - x}} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{1}{4\sqrt{x}\sqrt{\sin^{-1}(\sqrt{x})}\sqrt{1 - x}} \]