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.