To solve the problem of finding the derivative of the function $y = (5 + 15x) e^{-3x}$, we will apply the product rule of differentiation.

The product rule states: $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$ where $f(x) = (5 + 15x)$ and $g(x) = e^{-3x}$.

Step 1: Differentiate $f(x) = 5 + 15x$

$f'(x) = 15$

Step 2: Differentiate $g(x) = e^{-3x}$

The derivative of $e^{-3x}$ is: $g'(x) = \frac{d}{dx}[e^{-3x}] = -3e^{-3x}$

Step 3: Apply the product rule

Now we apply the product rule to find the derivative of $y$: $\frac{dy}{dx} = f'(x)g(x) + f(x)g'(x)$ Substituting the values of $f'(x)$, $f(x)$, $g(x)$, and $g'(x)$: $\frac{dy}{dx} = 15e^{-3x} + (5 + 15x)(-3e^{-3x})$

Step 4: Simplify the expression

Distribute the $-3e^{-3x}$ term: $\frac{dy}{dx} = 15e^{-3x} - 3(5 + 15x)e^{-3x}$ $\frac{dy}{dx} = 15e^{-3x} - 15e^{-3x} - 45xe^{-3x}$

Finally, simplifying further: $\frac{dy}{dx} = -45xe^{-3x}$

Thus, the derivative of $y$ with respect to $x$ is: $\boxed{\frac{dy}{dx} = -45xe^{-3x}}$

Would you like further details on any of these steps?

Here are some related questions to deepen your understanding:

1. How does the product rule differ from the chain rule in differentiation?
2. Can you explain why the derivative of $e^{-3x}$ includes the factor of $-3$?
3. How would you apply the quotient rule to a similar function with a division of terms?
4. What happens if you change the exponent in the exponential function from $-3x$ to $-kx$?
5. How would this derivative be applied in a real-world scenario, like calculating a rate of change?

Tip: Understanding when to use the product rule and when to use the chain rule is essential for differentiating more complex functions effectively!