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The given function is:

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

We will differentiate this using the **product rule**. The product rule states:

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

Here, \( f(x) = 1 + x^3 \) and \( g(x) = 3 - 2x \).

### Step 1: Find \( f'(x) \) and \( g'(x) \)

- \( f'(x) = \frac{d}{dx} (1 + x^3) = 3x^2 \)
- \( g'(x) = \frac{d}{dx} (3 - 2x) = -2 \)

### Step 2: Apply the product rule

Using the product rule, we get:

\[
y' = f'(x)g(x) + f(x)g'(x)
\]

Substitute the derivatives:

\[
y' = (3x^2)(3 - 2x) + (1 + x^3)(-2)
\]

### Step 3: Simplify the expression

First, expand both terms:

\[
y' = 3x^2(3 - 2x) + (-2)(1 + x^3)
\]

\[
y' = 9x^2 - 6x^3 - 2 - 2x^3
\]

Combine like terms:

\[
y' = -8x^3 + 9x^2 - 2
\]

Thus, the derivative of the function is:

\[
y' = -8x^3 + 9x^2 - 2
\]

Would you like any more details on this process?

Here are 5 related questions for further practice:

1. What is the second derivative of \( y = (1 + x^3)(3 - 2x) \)?
2. How does the product rule differ from the chain rule in differentiation?
3. What is the derivative of \( y = (1 + x^4)(2 - x) \)?
4. Can you find the critical points of the function by solving \( y' = 0 \)?
5. How do we apply the quotient rule to a function of the form \( y = \frac{f(x)}{g(x)} \)?

**Tip:** Always simplify expressions as much as possible after applying the differentiation rules to avoid mistakes in later calculations.

Learn how to find the derivative of the function y = (1 + x^3)(3 - 2x) using the product rule. This step-by-step solution covers key differentiation concepts and simplifies the final expression for advanced high school or early college students.

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