Answer
$${e^t}\left( {2{t^3} + 6{t^2}} \right) - 2{e^{ - t}}\left( {{t^2} - 2t - 1} \right) + 16{e^{2t}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{u}}\left( t \right) = 2{t^3}{\bf{i}} + \left( {{t^2} - 1} \right){\bf{j}} - 8{\bf{k}}{\text{ and }}{\bf{v}}\left( t \right) = {e^t}{\bf{i}} + 2{e^{ - t}}{\bf{j}} - {e^{2t}}{\bf{k}} \cr
& {\bf{u}}\left( t \right) \cdot {\bf{v}}\left( t \right) \cr
& {\text{Calculate the derivative using the dot product rule}} \cr
& \frac{d}{{dt}}\left( {{\bf{u}}\left( t \right) \cdot {\bf{v}}\left( t \right)} \right) = {\bf{u}}'\left( t \right) \cdot {\bf{v}}\left( t \right) + {\bf{u}}\left( t \right) \cdot {\bf{v}}'\left( t \right) \cr
& = \left( {6{t^2}{\bf{i}} + 2t{\bf{j}}} \right) \cdot \left( {{e^t}{\bf{i}} + 2{e^{ - t}}{\bf{j}} - {e^{2t}}{\bf{k}}} \right) \cr
& \,\,\,\, + \left( {2{t^3}{\bf{i}} + \left( {{t^2} - 1} \right){\bf{j}} - 8{\bf{k}}} \right) \cdot \left( {{e^t}{\bf{i}} - 2{e^{ - t}}{\bf{j}} - 2{e^{2t}}{\bf{k}}} \right) \cr
& {\text{Calculate the dot product and simplify}} \cr
& = 6{t^2}{e^t} + 4t{e^{ - t}} + 2{t^3}{e^t} - 2{e^{ - t}}\left( {{t^2} - 1} \right) + 16{e^{2t}} \cr
& {\text{Factoring}} \cr
& = \left( {6{t^2}{e^t} + 2{t^3}{e^t}} \right) + \left( {4t{e^{ - t}} - 2{e^{ - t}}\left( {{t^2} - 1} \right)} \right) + 16{e^{2t}} \cr
& = {e^t}\left( {2{t^3} + 6{t^2}} \right) - 2{e^{ - t}}\left( {{t^2} - 2t - 1} \right) + 16{e^{2t}} \cr} $$
