To find the derivative of $f(x) = x^2 (x - 2)^4$, we'll use the product rule, which states that for functions $u(x)$ and $v(x)$, the derivative of their product is:

$(fg)' = f'g + fg'$

In this case:

• Let $u(x) = x^2$
• Let $v(x) = (x - 2)^4$

Now let's find the derivatives of $u(x)$ and $v(x)$ separately, and then apply the product rule.

Step 1: Differentiate $u(x) = x^2$

$u'(x) = 2x$

Step 2: Differentiate $v(x) = (x - 2)^4$

Using the chain rule, we get: $v'(x) = 4(x - 2)^3 \cdot 1 = 4(x - 2)^3$

Step 3: Apply the Product Rule

Using $(uv)' = u'v + uv'$, we have: $f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

Substitute the values we found: $f'(x) = (2x)(x - 2)^4 + (x^2)(4(x - 2)^3)$

Step 4: Simplify

We can factor out $2x(x - 2)^3$ from each term: $f'(x) = 2x(x - 2)^3 \left( (x - 2) + 2x \right)$

Simplify inside the parentheses: $f'(x) = 2x(x - 2)^3 (3x - 2)$

Final Answer

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

Would you like further details or have any questions on this solution?

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Here are some follow-up questions to deepen understanding:

1. What is the second derivative of $f(x) = x^2(x-2)^4$?
2. How would you find the critical points of $f(x) = x^2(x-2)^4$?
3. What are the inflection points of $f(x) = x^2(x-2)^4$?
4. How does the function $f(x) = x^2(x-2)^4$ behave as $x \to \infty$ and $x \to -\infty$?
5. What is the behavior of $f(x) = x^2(x-2)^4$ near $x = 2$?

Tip: When using the product rule, always check if you can factor out terms to simplify the expression after differentiating.