If `f(x)=sqrt(sinx)," find "f^(')(x)`.

Answer

To find the derivative of the function \( f(x) = \sqrt{\sin x} \), we will use the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative is given by \( f'(g(x)) \cdot g'(x) \).

### Step-by-Step Solution:

1. **Identify the outer and inner functions**:
- Let \( u = \sin x \) (inner function).
- Then, \( f(x) = \sqrt{u} \) (outer function).

2. **Differentiate the outer function**:
- The derivative of \( \sqrt{u} \) with respect to \( u \) is:
\[ \frac{d}{du}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \]

3. **Differentiate the inner function**:
- The derivative of \( \sin x \) with respect to \( x \) is:
\[ \frac{d}{dx}(\sin x) = \cos x \]

4. **Apply the chain rule**:
- Now, using the chain rule:
\[ f'(x) = \frac{d}{du}(\sqrt{u}) \cdot \frac{d}{dx}(\sin x) \]
- Substitute \( u = \sin x \):
\[ f'(x) = \frac{1}{2\sqrt{\sin x}} \cdot \cos x \]

5. **Final expression**:
- Therefore, the derivative \( f'(x) \) is:
\[ f'(x) = \frac{\cos x}{2\sqrt{\sin x}} \]

### Final Answer:
\[ f'(x) = \frac{\cos x}{2\sqrt{\sin x}} \]