The function given is:

$h(x) = x^3 f(x)$

To find $h'(x)$ (the derivative of $h(x)$), we will use the product rule. The product rule states:

$\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) v(x) + u(x) v'(x)$

In this case, let:

• $u(x) = x^3$
• $v(x) = f(x)$

Step 1: Differentiate $u(x) = x^3$

$u'(x) = 3x^2$

Step 2: Differentiate $v(x) = f(x)$

$v'(x) = f'(x)$

Step 3: Apply the product rule

$h'(x) = u'(x) v(x) + u(x) v'(x)$ $h'(x) = 3x^2 f(x) + x^3 f'(x)$

So, the derivative of $h(x)$ is:

$h'(x) = 3x^2 f(x) + x^3 f'(x)$

Let me know if you'd like a further explanation or have more questions!

Here are 5 related questions to expand on this:

1. How would you differentiate $h(x) = x^n f(x)$ for any power $n$?
2. What happens if $f(x)$ is a constant? How does the derivative simplify?
3. How would you apply the product rule to three functions, e.g., $h(x) = x^2 f(x) g(x)$?
4. Can you use the quotient rule to differentiate $h(x) = \frac{f(x)}{g(x)}$?
5. What is the chain rule, and how does it apply to composite functions?

Tip: When differentiating products of functions, the product rule always helps break down the calculation into manageable pieces.