The derivative of `f(x) = (sqrt(x) + (1)/(sqrt(x)))^(2)` is

Options

A. `sqrt(x) + (1)/(sqrt(x))`
B. `2 ((sqrt(x) + 1)/(sqrt(x)))`
C. `1 - (1)/(x^(2))`
D. `1 + (1)/(x^(2))`

Answer

To find the derivative of the function \( f(x) = \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \), we can follow these steps:

### Step 1: Expand the function
First, we will expand the function using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \).

Let:
- \( a = \sqrt{x} \)
- \( b = \frac{1}{\sqrt{x}} \)

Then,
\[ f(x) = a^2 + 2ab + b^2 \]
Calculating each term:
- \( a^2 = (\sqrt{x})^2 = x \)
- \( b^2 = \left(\frac{1}{\sqrt{x}}\right)^2 = \frac{1}{x} \)
- \( 2ab = 2 \cdot \sqrt{x} \cdot \frac{1}{\sqrt{x}} = 2 \)

Putting it all together, we have:
\[ f(x) = x + \frac{1}{x} + 2 \]

### Step 2: Differentiate the function
Now we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( x + \frac{1}{x} + 2 \right) \]

Using the differentiation rules:
- The derivative of \( x \) is \( 1 \).
- The derivative of \( \frac{1}{x} \) can be found using the power rule: \( \frac{d}{dx} x^{-1} = -1 \cdot x^{-2} = -\frac{1}{x^2} \).
- The derivative of a constant (2) is \( 0 \).

Thus,
\[ f'(x) = 1 - \frac{1}{x^2} + 0 \]
This simplifies to:
\[ f'(x) = 1 - \frac{1}{x^2} \]

### Final Answer
The derivative of \( f(x) = \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \) is:
\[ f'(x) = 1 - \frac{1}{x^2} \]