Answer
$y'=-xe^{-3x}(3x-2)$
Work Step by Step
$y=x^{2}e^{-3x}$
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:
$...=-xe^{-3x}(3x-2)$
