The derivative of `f(x) = (x + (1)/(x))^(3)`

Options

A. `3 x^(2) + (3)/(x^(4)) - 3`
B. `3 x^(2) - (3)/(x^(4)) + 3 - (3)/(x^(2))`
C. `3x^(2) + (3)/(x^(4)) - 3 + (3)/(x^(2))`
D. `3x^(2) + (3)/(x^(4)) 3 + (3)/(x^(2))`

Answer

To find the derivative of the function \( f(x) = \left( x + \frac{1}{x} \right)^3 \), we can follow these steps:

### Step 1: Identify the function
We have the function:
\[ f(x) = \left( x + \frac{1}{x} \right)^3 \]

### Step 2: Use the chain rule
To differentiate \( f(x) \), we will use the chain rule. The chain rule states that if you have a composite function \( g(h(x)) \), then the derivative is:
\[ g'(h(x)) \cdot h'(x) \]
In our case, let \( u = x + \frac{1}{x} \). Then, \( f(x) = u^3 \).

### Step 3: Differentiate the outer function
First, differentiate the outer function \( f(u) = u^3 \):
\[ f'(u) = 3u^2 \]

### Step 4: Differentiate the inner function
Next, differentiate the inner function \( u = x + \frac{1}{x} \):
\[ u' = \frac{d}{dx}\left(x + \frac{1}{x}\right) = 1 - \frac{1}{x^2} \]

### Step 5: Apply the chain rule
Now, applying the chain rule:
\[ f'(x) = f'(u) \cdot u' = 3u^2 \cdot \left(1 - \frac{1}{x^2}\right) \]

### Step 6: Substitute back for \( u \)
Substituting back \( u = x + \frac{1}{x} \):
\[ f'(x) = 3\left(x + \frac{1}{x}\right)^2 \cdot \left(1 - \frac{1}{x^2}\right) \]

### Step 7: Simplify the expression
Now, we can simplify the expression:
1. First, calculate \( \left(x + \frac{1}{x}\right)^2 \):
\[ \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \]
2. Now substitute this back into the derivative:
\[ f'(x) = 3\left(x^2 + 2 + \frac{1}{x^2}\right) \cdot \left(1 - \frac{1}{x^2}\right) \]

### Step 8: Expand and simplify further
Now, expand the expression:
\[ f'(x) = 3\left(x^2 + 2 + \frac{1}{x^2} - \frac{x^2}{x^2} - \frac{2}{x^2} - \frac{1}{x^4}\right) \]
\[ = 3\left(x^2 + 2 - 1 + \frac{1}{x^2} - \frac{2}{x^2} - \frac{1}{x^4}\right) \]
\[ = 3\left(x^2 + 1 - \frac{1}{x^2} - \frac{1}{x^4}\right) \]

### Final Derivative
Thus, the derivative of \( f(x) \) is:
\[ f'(x) = 3\left(x^2 + 1 - \frac{1}{x^2} - \frac{1}{x^4}\right) \]