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 of differentiation.

### Step-by-Step Solution:

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

2. **Differentiate \( u \) and \( v \):**
- The derivative of \( u \) is \( u' = 1 \) (since the derivative of \( x \) is 1 and the derivative of a constant is 0).
- The derivative of \( v \) is \( v' = 2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}} \) (using the chain rule).

3. **Apply the quotient rule:**
The quotient rule states that if \( f(x) = \frac{u}{v} \), then:
\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]
Substituting in our values:
\[ f'(x) = \frac{(2\sqrt{x}) \cdot (1) - (x - 4) \cdot \left(\frac{1}{\sqrt{x}}\right)}{(2\sqrt{x})^2} \]

4. **Simplify the expression:**
- The denominator becomes \( (2\sqrt{x})^2 = 4x \).
- The numerator becomes:
\[ 2\sqrt{x} - \frac{x - 4}{\sqrt{x}} = 2\sqrt{x} - \frac{x}{\sqrt{x}} + \frac{4}{\sqrt{x}} = 2\sqrt{x} - \sqrt{x} + \frac{4}{\sqrt{x}} = \sqrt{x} + \frac{4}{\sqrt{x}} \]
Thus, we have:
\[ f'(x) = \frac{\sqrt{x} + \frac{4}{\sqrt{x}}}{4x} \]

5. **Combine the terms in the numerator:**
To combine \( \sqrt{x} \) and \( \frac{4}{\sqrt{x}} \), we can write:
\[ f'(x) = \frac{\frac{x + 4}{\sqrt{x}}}{4x} = \frac{x + 4}{4x\sqrt{x}} \]

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

### Final Answer:
Thus, \( f'(1) = \frac{5}{4} \).

---