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To find the derivative \( f'(x) \) of the function 

\[
f(x) = (x^2 + 1)^3 (x^3 - 1)^2,
\]

we need to apply the product rule and the chain rule. Let's proceed step by step.

### Step 1: Define Parts of the Function
Let
\[
u = (x^2 + 1)^3 \quad \text{and} \quad v = (x^3 - 1)^2.
\]

Then,
\[
f(x) = u \cdot v.
\]

### Step 2: Apply the Product Rule
The product rule states that:
\[
f'(x) = u'v + uv'.
\]

So, we need to find \( u' \) and \( v' \).

### Step 3: Differentiate \( u = (x^2 + 1)^3 \)
Using the chain rule, we have:
\[
u' = 3(x^2 + 1)^2 \cdot (2x) = 6x(x^2 + 1)^2.
\]

### Step 4: Differentiate \( v = (x^3 - 1)^2 \)
Similarly, using the chain rule:
\[
v' = 2(x^3 - 1) \cdot (3x^2) = 6x^2(x^3 - 1).
\]

### Step 5: Substitute Back into the Product Rule
Now we can substitute \( u \), \( u' \), \( v \), and \( v' \) into the product rule:
\[
f'(x) = u'v + uv'.
\]
Substitute the expressions:
\[
f'(x) = \left(6x(x^2 + 1)^2\right)(x^3 - 1)^2 + (x^2 + 1)^3 \left(6x^2(x^3 - 1)\right).
\]

### Final Answer
Thus, the derivative \( f'(x) \) is:
\[
f'(x) = 6x(x^2 + 1)^2(x^3 - 1)^2 + 6x^2(x^2 + 1)^3(x^3 - 1).
\]

Would you like further simplification or additional details on any steps? 

Here are some related questions to explore further:
1. How can the chain rule be applied to other complex functions?
2. What is the product rule, and how does it apply to derivatives involving multiple functions?
3. How can higher-order derivatives be found for similar functions?
4. How does simplification impact the understanding of the derivative function?
5. What are some practical applications of this derivative in calculus problems?

**Tip:** Always simplify terms where possible in derivatives to make it easier to interpret and solve further problems.

Risolvi la seguente derivata f(x) = (x^2 + 1)^3 (x^3 - 1)^2

Learn how to solve the derivative of the function f(x) = (x^2 + 1)^3 (x^3 - 1)^2 using the product and chain rules. This problem explores key calculus concepts and provides a step-by-step solution. Suitable for students in Grades 11-12.

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