Let's tackle this problem in two parts as requested:

1. Product Rule Method:

   The function given is: $f(x) = x^4 (x^9 - 8)$ To find $f'(x)$, we will apply the product rule, which states that: $(uv)' = u'v + uv'$ Here, $u = x^4$ and $v = (x^9 - 8)$. Applying the product rule: $f'(x) = \left( x^4 \right)' (x^9 - 8) + \left( x^9 - 8 \right)' x^4$ Now, calculate the derivatives: $\left( x^4 \right)' = 4x^3 \quad \text{and} \quad \left( x^9 - 8 \right)' = 9x^8$ So, $f'(x) = 4x^3 (x^9 - 8) + 9x^8 x^4$ Simplifying the second term: $f'(x) = 4x^3 (x^9 - 8) + 9x^{12}$ Therefore, the result using the product rule is: $f'(x) = 4x^3 (x^9 - 8) + 9x^{12}$

2. Simplifying First:

   Now, simplify the function before differentiating: $f(x) = x^4 (x^9 - 8) = x^{13} - 8x^4$ Differentiate directly: $f'(x) = \left( x^{13} - 8x^4 \right)' = 13x^{12} - 32x^3$

Thus, you get the following results:

• Product Rule Result: $f'(x) = 4x^3 (x^9 - 8) + 9x^{12}$
• Simplifying First Result: $f'(x) = 13x^{12} - 32x^3$

The final answer in the box for simplified $f(x)$ is: $f'(x) = 13x^{12} - 32x^3$

Would you like more details on the steps or have any other questions?

Here are 5 questions that expand on the topic:

1. How would you apply the product rule to more than two functions?
2. Can the product rule be extended to functions of several variables?
3. How does the chain rule differ from the product rule in differentiation?
4. How would implicit differentiation affect applying the product rule?
5. What are some practical examples of using the product rule in physics?

Tip: When simplifying a function before differentiating, always double-check the algebra to avoid errors during differentiation.