To find the derivative of \( f(x) = \left( \frac{x - 7}{x + 2} \right)^2 \), we'll use the **chain rule** and the **quotient rule**.

### Step 1: Rewrite the function
Let \( g(x) = \frac{x - 7}{x + 2} \), so that \( f(x) = \left( g(x) \right)^2 \).

Now, apply the chain rule:
\[
f'(x) = 2 \cdot g(x) \cdot g'(x)
\]
So we need to find \( g'(x) \).

### Step 2: Use the quotient rule to find \( g'(x) \)
Recall the quotient rule:
\[
\left( \frac{u}{v} \right)' = \frac{v u' - u v'}{v^2}
\]
For \( g(x) = \frac{x - 7}{x + 2} \), let:
- \( u = x - 7 \)
- \( v = x + 2 \)

Now, differentiate both \( u \) and \( v \):
- \( u' = 1 \)
- \( v' = 1 \)

Using the quotient rule:
\[
g'(x) = \frac{(x + 2)(1) - (x - 7)(1)}{(x + 2)^2} = \frac{x + 2 - (x - 7)}{(x + 2)^2} = \frac{x + 2 - x + 7}{(x + 2)^2} = \frac{9}{(x + 2)^2}
\]

### Step 3: Substitute \( g(x) \) and \( g'(x) \) into the chain rule formula
Now that we have \( g'(x) = \frac{9}{(x + 2)^2} \) and \( g(x) = \frac{x - 7}{x + 2} \), we substitute into the chain rule:
\[
f'(x) = 2 \cdot \frac{x - 7}{x + 2} \cdot \frac{9}{(x + 2)^2}
\]
Simplifying:
\[
f'(x) = \frac{18(x - 7)}{(x + 2)^3}
\]

Thus, the derivative of \( f(x) = \left( \frac{x - 7}{x + 2} \right)^2 \) is:
\[
f'(x) = \frac{18(x - 7)}{(x + 2)^3}
\]

Would you like more details on any step?

### 5 Related Questions:
1. How does the chain rule work in more complex functions?
2. Can the quotient rule be used for functions with more than two terms?
3. What happens if we differentiate higher powers of \( g(x) \)?
4. How would you find the second derivative of \( f(x) \)?
5. How can we apply this process to functions with trigonometric terms?

**Tip**: When using the quotient rule, always simplify the result before applying it in further calculations to avoid errors.

Learn how to find the derivative of f(x) = (x - 7 / x + 2)^2 using the chain rule and the quotient rule. This detailed step-by-step solution covers key calculus concepts and is suitable for students studying undergraduate calculus or advanced high school calculus.

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