To find the derivative of $f(x) = x^2(x-2)^4$, we will apply both the product rule and the chain rule.

The product rule states that if $f(x) = u(x) \cdot v(x)$, then:

$f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

Step 1: Define $u(x)$ and $v(x)$

Let:

• $u(x) = x^2$
• $v(x) = (x-2)^4$

Now, we need to find the derivatives of $u(x)$ and $v(x)$.

Step 2: Find $u'(x)$

The derivative of $u(x) = x^2$ is:

$u'(x) = 2x$

Step 3: Find $v'(x)$ using the chain rule

The derivative of $v(x) = (x-2)^4$ can be found using the chain rule. The chain rule states that if $f(g(x))$, then:

$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

For $v(x) = (x-2)^4$:

• The outer function is $f(u) = u^4$, whose derivative is $4u^3$.
• The inner function is $g(x) = x-2$, whose derivative is $1$.

Thus, applying the chain rule:

$v'(x) = 4(x-2)^3 \cdot 1 = 4(x-2)^3$

Step 4: Apply the product rule

Now, using the product rule:

$f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

Substitute the values we found:

$f'(x) = 2x \cdot (x-2)^4 + x^2 \cdot 4(x-2)^3$

Step 5: Simplify the expression

Factor out common terms from both parts of the sum:

$f'(x) = 2x(x-2)^3 \left[ (x-2) + 2x \right]$

Simplify the terms inside the brackets:

$f'(x) = 2x(x-2)^3 (3x - 2)$

This is the derivative of $f(x) = x^2(x-2)^4$.

Final answer:

$f'(x) = 2x(x-2)^3(3x-2)$

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Would you like further clarification on any part of the process? Here are some related questions you might consider:

1. How is the chain rule used in more complex functions?
2. What is the product rule, and how can it be generalized for more than two factors?
3. Can this method be applied to other polynomial expressions?
4. What if we wanted to find the second derivative of this function?
5. How can we use this derivative to find critical points and analyze the graph?

Tip: Always check if factoring simplifies the derivative before expanding—this often reduces the complexity of the final answer.