Answer
$y'=\dfrac{1-x}{e^{x}}$
Work Step by Step
$y=\dfrac{x}{e^{x}}$
Differentiate using the quotient rule:
$y'=\dfrac{(e^{x})(x)'-(x)(e^{x})'}{(e^{x})^{2}}=\dfrac{e^{x}-xe^{x}}{e^{2x}}$
Take out common factor $e^{x}$
$y'=\dfrac{e^{x}(1-x)}{e^{2x}}=\dfrac{1-x}{e^{x}}$
$y'=\dfrac{1-x}{e^{x}}$