Answer
$y'=e^{-1/x}(1+2x)$
Work Step by Step
$y=x^{2}e^{-1/x}$
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:
$...=e^{-1/x}+2xe^{-1/x}=e^{-1/x}(1+2x)$
