Answer
\[\left( {8{t^3} - 4t} \right){\mathbf{i}} - \left( { - 6{t^5} + 2} \right){\mathbf{j}} + \left( { - 5{t^4} + 4{t^3}} \right){\mathbf{k}}\]
Work Step by Step
\[\begin{gathered}
\frac{d}{{dt}}\left( {\left( {{t^3}{\mathbf{i}} - 2t{\mathbf{j}} - 2{\mathbf{k}}} \right) \times \left( {t{\mathbf{i}} - {t^2}{\mathbf{j}} - {t^3}{\mathbf{k}}} \right)} \right) \hfill \\
\hfill \\
\left( {{t^3}{\mathbf{i}} - 2t{\mathbf{j}} - 2{\mathbf{k}}} \right) \times \left( {t{\mathbf{i}} - {t^2}{\mathbf{j}} - {t^3}{\mathbf{k}}} \right) \hfill \\
= \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{{t^3}}&{ - 2t}&{ - 2} \\
t&{ - {t^2}}&{ - {t^3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{ - 2t}&{ - 2} \\
{ - {t^2}}&{ - {t^3}}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{{t^3}}&{ - 2} \\
t&{ - {t^3}}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{{t^3}}&{ - 2t} \\
t&{ - {t^2}}
\end{array}} \right|{\mathbf{k}} \hfill \\
= \left( {2{t^4} - 2{t^2}} \right){\mathbf{i}} - \left( { - {t^6} + 2t} \right){\mathbf{j}} + \left( { - {t^5} + 2{t^2}} \right){\mathbf{k}} \hfill \\
\hfill \\
{\text{Calculate the derivative}} \hfill \\
= \left( {8{t^3} - 4t} \right){\mathbf{i}} - \left( { - 6{t^5} + 2} \right){\mathbf{j}} + \left( { - 5{t^4} + 4{t^3}} \right){\mathbf{k}} \hfill \\
\end{gathered} \]
