To find the derivative of the function $f(x) = (3x + 3)e^x$, we will use the product rule. The product rule states that if $u(x)$ and $v(x)$ are functions of $x$, then the derivative $(uv)' = u'v + uv'$.

Let:

• $u = 3x + 3$
• $v = e^x$

First, we calculate the derivatives:

• $u' = 3$
• $v' = e^x$

Now apply the product rule: $f'(x) = u'v + uv' = (3)(e^x) + (3x + 3)(e^x)$

Factor out $e^x$: $f'(x) = e^x(3 + (3x + 3)) = e^x(3x + 6)$

Thus, the derivative is: $f'(x) = e^x(3x + 6)$

Would you like more details on this process or have any specific questions? Here are some related questions you might find interesting:

1. What is the second derivative of the function?
2. How can we evaluate the derivative at a specific point, say $x = 1$?
3. What is the behavior of $f(x)$ as $x \to \infty$?
4. How does the product rule compare to the quotient rule?
5. Can you explain how to find critical points using the derivative?

Tip: Always remember to check the conditions for applying the product rule or any other differentiation rules.