## Calculus: Early Transcendentals 8th Edition

$y'=e^{-1/x}(1+2x)$
$y=x^{2}e^{-1/x}$ Differentiate using the product rule: $y'=(x^{2})(e^{-1/x})'+(e^{-1/x})(x^{2})'=...$ Use the chain rule to find $(e^{-1/x})'$: $...=(x^{2})[(e^{-1/x})(-\dfrac{1}{x})']+(e^{-1/x})(2x)=...$ Write $-\dfrac{1}{x}$ as $-x^{-1}$ and continue with the differentiation process: $...=(x^{2})[(e^{-1/x})(-x^{-1})']+(e^{-1/x})(2x)=...$ $...=(x^{2})[(e^{-1/x})(x^{-2})]+(e^{-1/x})(2x)=...$ Simplify: $...=e^{-1/x}+2xe^{-1/x}=e^{-1/x}(1+2x)$