Answer
$V'\left( t \right) = \frac{{3t + 2{e^t} + 4t{e^t}}}{{2\sqrt t }}$
Work Step by Step
$$\eqalign{
& V\left( t \right) = \left( {t + 2{e^t}} \right)\sqrt t \cr
& {\text{Differentiating}} \cr
& V'\left( t \right) = \frac{d}{{dt}}\left[ {\left( {t + 2{e^t}} \right)\sqrt t } \right] \cr
& {\text{Differentiate by using the formula }}\left( {gh} \right)' = gh' + hg' \cr
& {\text{Let }}g = t + 2{e^t}{\text{ and }}h = \sqrt t ,{\text{ then}} \cr
& V'\left( t \right) = \left( {t + 2{e^t}} \right)\left( {\sqrt t } \right)' + \sqrt t \left( {t + 2{e^t}} \right)' \cr
& {\text{Compute derivatives}} \cr
& V'\left( t \right) = \left( {t + 2{e^t}} \right)\left( {\frac{1}{{2\sqrt t }}} \right) + \sqrt t \left( {1 + 2{e^t}} \right) \cr
& {\text{Multiply and simplify}} \cr
& V'\left( t \right) = \frac{t}{{2\sqrt t }} + \frac{{2{e^t}}}{{2\sqrt t }} + \sqrt t + 2\sqrt t {e^t} \cr
& V'\left( t \right) = \frac{{t + 2{e^t} + 2\left( t \right) + 2\left( t \right){e^t}}}{{2\sqrt t }} \cr
& V'\left( t \right) = \frac{{3t + 2{e^t} + 4t{e^t}}}{{2\sqrt t }} \cr} $$
