Answer
$g'(x)=e^{x}(x+2\sqrt{x}+1+\dfrac{1}{\sqrt{x}})$
Work Step by Step
$g(x)=(x+2\sqrt{x})e^{x}$
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:
$g'(x)=e^{x}(x+2\sqrt{x}+1+\dfrac{1}{\sqrt{x}})$
