If `f(x)=sqrt(ax)+(a^2)/(sqrt(ax))`, then `f^`(a)=`

Options

A. -1
B. 1
C. 0
D. a

Explanation

To solve the problem, we need to find the derivative of the function \( f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}} \) and then evaluate it at \( x = a \).

### Step-by-step Solution:

1. **Rewrite the Function**:
We start with the function:
\[ f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}} \]
We can express \( \sqrt{ax} \) as \( \sqrt{a} \cdot \sqrt{x} \) and \( \frac{a^2}{\sqrt{ax}} \) as \( a^2 \cdot \frac{1}{\sqrt{ax}} = a^2 \cdot \frac{1}{\sqrt{a} \cdot \sqrt{x}} = a^{3/2} \cdot x^{-1/2} \).

2. **Differentiate \( f(x) \)**:
The function can now be rewritten as:
\[ f(x) = \sqrt{a} \cdot x^{1/2} + a^{3/2} \cdot x^{-1/2} \]
We will differentiate \( f(x) \) using the power rule:
\[ f'(x) = \frac{d}{dx}(\sqrt{a} \cdot x^{1/2}) + \frac{d}{dx}(a^{3/2} \cdot x^{-1/2}) \]
- The derivative of \( \sqrt{a} \cdot x^{1/2} \) is:
\[ \sqrt{a} \cdot \frac{1}{2} x^{-1/2} = \frac{\sqrt{a}}{2\sqrt{x}} \]
- The derivative of \( a^{3/2} \cdot x^{-1/2} \) is:
\[ a^{3/2} \cdot \left(-\frac{1}{2} x^{-3/2}\right) = -\frac{a^{3/2}}{2x^{3/2}} \]
Therefore, combining these results, we have:
\[ f'(x) = \frac{\sqrt{a}}{2\sqrt{x}} - \frac{a^{3/2}}{2x^{3/2}} \]

3. **Evaluate \( f'(a) \)**:
Now we substitute \( x = a \) into \( f'(x) \):
\[ f'(a) = \frac{\sqrt{a}}{2\sqrt{a}} - \frac{a^{3/2}}{2a^{3/2}} \]
Simplifying this gives:
\[ f'(a) = \frac{1}{2} - \frac{1}{2} = 0 \]

### Final Answer:
Thus, the value of \( f'(a) \) is:
\[ \boxed{0} \]