We are tasked with finding the derivative of:

$y = (3x^2 - 6x + 6)e^{-5x}$

This is a product of two functions: $u(x) = 3x^2 - 6x + 6$ and $v(x) = e^{-5x}$. To differentiate this, we will apply the product rule:

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

Step 1: Differentiate $u(x)$

$u(x) = 3x^2 - 6x + 6$ $u'(x) = \frac{d}{dx}(3x^2 - 6x + 6) = 6x - 6$

Step 2: Differentiate $v(x)$

$v(x) = e^{-5x}$ $v'(x) = \frac{d}{dx}(e^{-5x}) = -5e^{-5x}$

Step 3: Apply the product rule

Now, applying the product rule:

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

Substituting the derivatives:

$\frac{dy}{dx} = (6x - 6)e^{-5x} + (3x^2 - 6x + 6)(-5e^{-5x})$

Step 4: Simplify the expression

Factor out $e^{-5x}$:

$\frac{dy}{dx} = e^{-5x} \left[ (6x - 6) + (-5)(3x^2 - 6x + 6) \right]$

Expand the second term:

$= e^{-5x} \left[ 6x - 6 - 15x^2 + 30x - 30 \right]$

Combine like terms:

$= e^{-5x} \left[ -15x^2 + 36x - 36 \right]$

Thus, the derivative is:

$\frac{dy}{dx} = e^{-5x}(-15x^2 + 36x - 36)$

Let me know if you need further details or have any questions!

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Here are 5 related questions:

1. How does the product rule apply to other exponential functions?
2. What is the significance of negative exponents in derivatives?
3. How do you factor a quadratic function like $-15x^2 + 36x - 36$?
4. How can you check if the derivative is correct using limits?
5. What are some real-world applications of exponential derivatives?

Tip: When using the product rule, always differentiate each function separately before combining them!