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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!

Find the derivative of y with respect to x. y = (5 + 15x)e^(-3x)

Learn how to find the derivative of the function y = (5 + 15x)e^(-3x) using the product rule of differentiation. This problem involves exponential functions and is suitable for students in Grades 11-12 or early college.

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