## Calculus: Early Transcendentals 8th Edition

$y'=-xe^{-3x}(3x-2)$
$y=x^{2}e^{-3x}$ Differentiate using the product rule: $y'=(x^{2})(e^{-3x})'+(e^{-3x})(x^{2})'=...$ Now, apply the chain rule to find $(e^{-3x})'$: $...=(x^{2})[(e^{-3x})(-3x)']+(e^{-3x})(2x)=(x^{2})(-3e^{-3x})+2xe^{-3x}=...$ $...=-3x^{2}e^{-3x}+2xe^{-3x}=...$ Take out common factor $-x^{2}e^{-3x}$ to present a better looking answer: $...=-xe^{-3x}(3x-2)$