To determine the derivative [tex]\( f'(x) \)[/tex] of the function [tex]\( f(x) = \left( x + \frac{1}{x} \right)^2 \)[/tex], let's proceed step by step.
### Step 1: Write the Function
Given:
[tex]\[ f(x) = \left( x + \frac{1}{x} \right)^2 \][/tex]
### Step 2: Apply the Chain Rule
The chain rule states that if you have a composite function [tex]\( g(h(x)) \)[/tex], then the derivative [tex]\( g(h(x))' = g'(h(x)) \cdot h'(x) \)[/tex]. Identify the outer function and the inner function:
- Outer function: [tex]\( u^2 \)[/tex] where [tex]\( u = x + \frac{1}{x} \)[/tex]
- Inner function: [tex]\( u = x + \frac{1}{x} \)[/tex]
### Step 3: Differentiate the Outer Function
The derivative of [tex]\( u^2 \)[/tex] with respect to [tex]\( u \)[/tex] is:
[tex]\[ \frac{d}{du}(u^2) = 2u \][/tex]
### Step 4: Differentiate the Inner Function
The inner function [tex]\( u \)[/tex] is:
[tex]\[ u = x + \frac{1}{x} \][/tex]
Taking the derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} \left( x + \frac{1}{x} \right) = 1 - \frac{1}{x^2} \][/tex]
### Step 5: Apply the Chain Rule
Combine the results from steps 3 and 4:
[tex]\[ f'(x) = \frac{d}{dx}\left( \left( x + \frac{1}{x} \right)^2 \right) = 2\left( x + \frac{1}{x} \right) \cdot \left( 1 - \frac{1}{x^2} \right) \][/tex]
### Step 6: Simplify the Expression
Finally, simplify the expression:
[tex]\[ f'(x) = 2\left( x + \frac{1}{x} \right)\left( 1 - \frac{1}{x^2} \right) \][/tex]
Therefore, the derivative [tex]\( f'(x) \)[/tex] is:
[tex]\[ f'(x) = \left( 2 - \frac{2}{x^2} \right)\left( x + \frac{1}{x} \right) \][/tex]
