Answer
$W'\left( t \right) = \left( {1 + 2t{e^t} + {e^t}} \right){e^t}$
Work Step by Step
$$\eqalign{
& W\left( t \right) = {e^t}\left( {1 + t{e^t}} \right) \cr
& {\text{Distribute}} \cr
& W\left( t \right) = {e^t}\left( 1 \right) + {e^t}\left( {t{e^t}} \right) \cr
& W\left( t \right) = {e^t} + t{e^{2t}} \cr
& {\text{Differentiating}} \cr
& W'\left( t \right) = \frac{d}{{dt}}\left[ {{e^t} + t{e^{2t}}} \right] \cr
& {\text{Sum rule for derivatives}} \cr
& W'\left( t \right) = \frac{d}{{dt}}\left[ {{e^t}} \right] + \frac{d}{{dt}}\left[ {t{e^{2t}}} \right] \cr
& {\text{Differentiate }}\frac{d}{{dt}}\left[ {t{e^{2t}}} \right]\,{\text{ by using the formula }}\left( {fg} \right)' = fg' + gf' \cr
& W'\left( t \right) = \frac{d}{{dt}}\left[ {{e^t}} \right] + t\frac{d}{{dt}}\left[ {{e^{2t}}} \right] + {e^{2t}}\frac{d}{{dt}}\left[ t \right] \cr
& {\text{Computing derivatives}} \cr
& W'\left( t \right) = {e^t} + t\left( {2{e^{2t}}} \right) + {e^{2t}}\left( 1 \right) \cr
& {\text{Multiply and simplify}} \cr
& W'\left( t \right) = {e^t} + 2t{e^{2t}} + {e^{2t}} \cr
& {\text{Factoring}} \cr
& W'\left( t \right) = \left( {1 + 2t{e^t} + {e^t}} \right){e^t} \cr} $$
