Answer
$g'\left( x \right) = \left( {x + 2\sqrt x + 1 + \frac{1}{{\sqrt x }}} \right){e^x}$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \left( {x + 2\sqrt x } \right){e^x} \cr
& {\text{Differentiating}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\left( {x + 2\sqrt x } \right){e^x}} \right] \cr
& {\text{Differentiate by using the formula }}\left( {fh} \right)' = fh' + hf' \cr
& {\text{Let }}f = x + 2\sqrt x {\text{ and }}h = {e^x},{\text{ then}} \cr
& g'\left( x \right) = \left( {x + 2\sqrt x } \right)\left( {{e^x}} \right)' + {e^x}\left( {x + 2\sqrt x } \right)' \cr
& {\text{Compute derivatives}} \cr
& g'\left( x \right) = \left( {x + 2\sqrt x } \right)\left( {{e^x}} \right) + {e^x}\left( {1 + 2\left( {\frac{1}{{2\sqrt x }}} \right)} \right) \cr
& {\text{Multiply}} \cr
& g'\left( x \right) = \left( {x + 2\sqrt x } \right)\left( {{e^x}} \right) + {e^x}\left( {1 + \frac{1}{{\sqrt x }}} \right) \cr
& {\text{Factoring}} \cr
& g'\left( x \right) = \left( {x + 2\sqrt x + 1 + \frac{1}{{\sqrt x }}} \right){e^x} \cr} $$
