## Calculus: Early Transcendentals 8th Edition

$f'(x)=e^{x}(3x^{2}+x-5)$
$f(x)=(3x^{2}-5x)e^{x}$ Differentiate by applying the product rule: $f'(x)=(3x^{2}-5x)(e^{x})'+(e^{x})(3x^{2}-5x)'=...$ $...=(3x^{2}-5x)(e^{x})+(e^{x})(6x-5)=...$ Take out common factor $e^{x}$: $...=e^{x}(3x^{2}-5x+6x-5)=e^{x}(3x^{2}+x-5)$ $f'(x)=e^{x}(3x^{2}+x-5)$