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

Options

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

Explanation

To find \( f'(a) \) for the function \( f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}} \), we will follow these steps:

### Step 1: Differentiate \( f(x) \)

We start with the function:
\[ f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}} \]

We will differentiate \( f(x) \) with respect to \( x \).

### Step 2: Differentiate \( \sqrt{ax} \)

Using the chain rule, the derivative of \( \sqrt{ax} \) is:
\[ \frac{d}{dx}(\sqrt{ax}) = \frac{1}{2\sqrt{ax}} \cdot a = \frac{a}{2\sqrt{ax}} \]

### Step 3: Differentiate \( \frac{a^2}{\sqrt{ax}} \)

Using the quotient rule, we differentiate \( \frac{a^2}{\sqrt{ax}} \):
\[ \frac{d}{dx}\left(\frac{a^2}{\sqrt{ax}}\right) = a^2 \cdot \frac{d}{dx}(\sqrt{ax})^{-1} \]
Using the chain rule:
\[ \frac{d}{dx}(\sqrt{ax})^{-1} = -\frac{1}{(\sqrt{ax})^2} \cdot \frac{d}{dx}(\sqrt{ax}) = -\frac{1}{ax} \cdot \frac{a}{2\sqrt{ax}} = -\frac{a}{2ax\sqrt{ax}} \]
Thus,
\[ \frac{d}{dx}\left(\frac{a^2}{\sqrt{ax}}\right) = a^2 \cdot -\frac{a}{2ax\sqrt{ax}} = -\frac{a^2}{2x\sqrt{ax}} \]

### Step 4: Combine the derivatives

Now we combine the derivatives:
\[ f'(x) = \frac{a}{2\sqrt{ax}} - \frac{a^2}{2x\sqrt{ax}} \]

### Step 5: Simplify \( f'(x) \)

We can factor out \( \frac{1}{2\sqrt{ax}} \):
\[ f'(x) = \frac{1}{2\sqrt{ax}} \left( a - \frac{a^2}{x} \right) \]

### Step 6: Evaluate \( f'(a) \)

Now we substitute \( x = a \):
\[ f'(a) = \frac{1}{2\sqrt{a^2}} \left( a - \frac{a^2}{a} \right) = \frac{1}{2a} \left( a - a \right) = \frac{1}{2a} \cdot 0 = 0 \]

### Final Answer

Thus, the value of \( f'(a) \) is:
\[ f'(a) = 0 \]