## Calculus: Early Transcendentals 8th Edition

$g'(x)=e^{x}(x+2\sqrt{x}+1+\dfrac{1}{\sqrt{x}})$
$g(x)=(x+2\sqrt{x})e^{x}$ Write the function in the following form: $g(x)=(x+2x^{1/2})e^{x}$ Differentiate by applying the product rule: $g'(x)=(x+2x^{1/2})(e^{x})'+(e^{x})(x+2x^{1/2})'=...$ $...=(x+2x^{1/2})(e^{x})+(e^{x})(1+2(\dfrac{1}{2})x^{-1/2})=...$ $...=(x+2x^{1/2})(e^{x})+(e^{x})(1+x^{-1/2})$ Take out common factor $e^{x}$: $g'(x)=e^{x}(x+2x^{1/2}+1+x^{-1/2})$ Rewrite like this for a better looking answer: $g'(x)=e^{x}(x+2\sqrt{x}+1+\dfrac{1}{\sqrt{x}})$