To find the derivative of the function $y = (1 + 6x)e^{-6x}$, we will use the product rule because this is a product of two functions: $(1 + 6x)$ and $e^{-6x}$.

The product rule states: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$ where $u(x) = 1 + 6x$ and $v(x) = e^{-6x}$.

Step 1: Differentiate $u(x) = 1 + 6x$

$u'(x) = \frac{d}{dx}(1 + 6x) = 6$

Step 2: Differentiate $v(x) = e^{-6x}$

To differentiate $e^{-6x}$, we apply the chain rule. The derivative of $e^{g(x)}$ is $g'(x)e^{g(x)}$, where $g(x) = -6x$.

$v'(x) = \frac{d}{dx}(e^{-6x}) = -6e^{-6x}$

Step 3: Apply the product rule

Now, using the product rule: $y' = u'(x)v(x) + u(x)v'(x)$ $y' = (6)(e^{-6x}) + (1 + 6x)(-6e^{-6x})$ $y' = 6e^{-6x} - 6(1 + 6x)e^{-6x}$ $y' = 6e^{-6x} - 6e^{-6x} - 36xe^{-6x}$ $y' = -36xe^{-6x}$

Final answer:

$\boxed{y' = -36xe^{-6x}}$

Do you need further clarification or any specific details on this?

Here are some follow-up questions:

1. How do you apply the chain rule when differentiating exponential functions?
2. Can you explain the product rule in more depth with another example?
3. How do you differentiate a function involving both an exponential and a polynomial?
4. What is the meaning of $e^{-6x}$ in terms of exponential decay?
5. How do you approach differentiation of more complex functions with multiple terms?

Tip: When differentiating products of functions, always remember the product rule—it's crucial when you have multiple parts of a function multiplied together.