Let `y= (x + (1)/(x))^(2)`Find the rate of change of y with respect to x when x=2.

Answer

To find the rate of change of \( y \) with respect to \( x \) when \( x = 2 \) for the function \( y = \left( x + \frac{1}{x} \right)^2 \), we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)
Given:
\[ y = \left( x + \frac{1}{x} \right)^2 \]
We will use the chain rule to differentiate this expression. Let \( u = x + \frac{1}{x} \), then \( y = u^2 \).

Using the chain rule:
\[ \frac{dy}{dx} = 2u \cdot \frac{du}{dx} \]

### Step 2: Find \( \frac{du}{dx} \)
Now we need to differentiate \( u \):
\[ u = x + \frac{1}{x} \]
Differentiating \( u \):
\[ \frac{du}{dx} = 1 - \frac{1}{x^2} \]

### Step 3: Substitute \( u \) and \( \frac{du}{dx} \) back into the derivative of \( y \)
Substituting \( u \) and \( \frac{du}{dx} \) into the derivative of \( y \):
\[ \frac{dy}{dx} = 2 \left( x + \frac{1}{x} \right) \left( 1 - \frac{1}{x^2} \right) \]

### Step 4: Simplify the expression
Now we simplify:
\[ \frac{dy}{dx} = 2 \left( x + \frac{1}{x} \right) \left( 1 - \frac{1}{x^2} \right) \]
\[ = 2 \left( x + \frac{1}{x} \right) \left( \frac{x^2 - 1}{x^2} \right) \]
\[ = 2 \cdot \frac{(x + \frac{1}{x})(x^2 - 1)}{x^2} \]

### Step 5: Evaluate \( \frac{dy}{dx} \) at \( x = 2 \)
Now we substitute \( x = 2 \):
\[ \frac{dy}{dx} = 2 \cdot \frac{\left( 2 + \frac{1}{2} \right) \left( 2^2 - 1 \right)}{2^2} \]
Calculating:
\[ = 2 \cdot \frac{\left( 2 + 0.5 \right) \left( 4 - 1 \right)}{4} \]
\[ = 2 \cdot \frac{2.5 \cdot 3}{4} \]
\[ = 2 \cdot \frac{7.5}{4} \]
\[ = \frac{15}{4} \]

### Final Answer
The rate of change of \( y \) with respect to \( x \) when \( x = 2 \) is:
\[ \frac{dy}{dx} = \frac{15}{4} \]