To find the derivative of the function $y = (1 + 9x)e^{-9x}$, we will use the product rule for differentiation. The product rule states:

$\frac{d}{dx} \left[ u(x) \cdot v(x) \right] = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

In this case, we define:

• $u(x) = 1 + 9x$
• $v(x) = e^{-9x}$

Step 1: Differentiate $u(x)$

The derivative of $u(x) = 1 + 9x$ is:

$u'(x) = 9$

Step 2: Differentiate $v(x)$

The derivative of $v(x) = e^{-9x}$ requires the chain rule. The derivative of $e^{-9x}$ is:

$v'(x) = e^{-9x} \cdot (-9) = -9e^{-9x}$

Step 3: Apply the product rule

Now, applying the product rule:

$\frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

Substitute the values we found:

$\frac{dy}{dx} = 9 \cdot e^{-9x} + (1 + 9x) \cdot (-9e^{-9x})$

Simplify the expression:

$\frac{dy}{dx} = 9e^{-9x} - 9(1 + 9x)e^{-9x}$

Factor out $e^{-9x}$:

$\frac{dy}{dx} = e^{-9x} \left( 9 - 9(1 + 9x) \right)$

Simplify the terms inside the parentheses:

$\frac{dy}{dx} = e^{-9x} \left( 9 - 9 - 81x \right)$

$\frac{dy}{dx} = e^{-9x} \left( -81x \right)$

Thus, the derivative of the function is:

$\frac{dy}{dx} = -81x e^{-9x}$

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Would you like further details on this solution or have any questions?

Here are some related questions you might find interesting:

1. How do you apply the chain rule when differentiating exponential functions?
2. Can you apply the product rule to other types of functions, like trigonometric ones?
3. What is the general formula for differentiating products of two functions?
4. How do you simplify derivatives that contain common terms?
5. How would you compute the second derivative of this function?

Tip: Always double-check the application of the chain and product rules when handling complex functions.