If f(x) `=(x-4)/(2sqrt(x))`, then `f^(')(1)` is equal to

Options

A. `5/4`
B. `4/5`
C. 1
D. 0

Explanation

To find \( f'(1) \) for the function \( f(x) = \frac{x - 4}{2\sqrt{x}} \), we will use the quotient rule for differentiation. The quotient rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

### Step-by-step Solution:

1. **Identify \( u \) and \( v \)**:
- Let \( u = x - 4 \)
- Let \( v = 2\sqrt{x} \)

2. **Calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)**:
- \( \frac{du}{dx} = 1 \) (derivative of \( x - 4 \))
- \( \frac{dv}{dx} = \frac{d}{dx}(2\sqrt{x}) = 2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}} \)

3. **Apply the Quotient Rule**:
\[ f'(x) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ f'(x) = \frac{2\sqrt{x} \cdot 1 - (x - 4) \cdot \frac{1}{\sqrt{x}}}{(2\sqrt{x})^2} \]

4. **Simplify the expression**:
\[ f'(x) = \frac{2\sqrt{x} - \frac{x - 4}{\sqrt{x}}}{4x} \]
To combine the terms in the numerator:
\[ f'(x) = \frac{2\sqrt{x} - \frac{x - 4}{\sqrt{x}}}{4x} = \frac{2x - (x - 4)}{4x\sqrt{x}} = \frac{2x - x + 4}{4x\sqrt{x}} = \frac{x + 4}{4x\sqrt{x}} \]

5. **Evaluate \( f'(1) \)**:
Substitute \( x = 1 \):
\[ f'(1) = \frac{1 + 4}{4 \cdot 1 \cdot \sqrt{1}} = \frac{5}{4} \]

### Final Answer:
\[ f'(1) = \frac{5}{4} \]

---