Answer
\[\left( {72{t^{ - 3}} - 60{t^{3/2}}} \right){\mathbf{i}} - \left( { - 6 + 9{t^{1/2}}} \right){\mathbf{j}} + \left( { - 60{t^4} - 18} \right){\mathbf{k}}\]
Work Step by Step
\[\begin{gathered}
\frac{d}{{dt}}\left( {\left( {{t^3}{\mathbf{i}} + 6{\mathbf{j}} - 2\sqrt t {\mathbf{k}}} \right) \cdot \left( {3t{\mathbf{i}} - 12{t^2}{\mathbf{j}} - 6{t^{ - 2}}{\mathbf{k}}} \right)} \right) \hfill \\
{\text{Calculate the cross product}} \hfill \\
= \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{{t^3}}&6&{ - 2\sqrt t } \\
{3t}&{ - 12{t^2}}&{ - 6{t^{ - 2}}}
\end{array}} \right| \hfill \\
= \left| {\begin{array}{*{20}{c}}
6&{ - 2\sqrt t } \\
{ - 12{t^2}}&{ - 6{t^{ - 2}}}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{{t^3}}&{ - 2\sqrt t } \\
{3t}&{ - 6{t^{ - 2}}}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{{t^3}}&6 \\
{3t}&{ - 12{t^2}}
\end{array}} \right|{\mathbf{k}} \hfill \\
= \left( { - 36{t^{ - 2}} - 24{t^{5/2}}} \right){\mathbf{i}} - \left( { - 6t + 6{t^{3/2}}} \right){\mathbf{j}} + \left( { - 12{t^5} - 18t} \right){\mathbf{k}} \hfill \\
{\text{Calculate the derivative}} \hfill \\
= \left( {72{t^{ - 3}} - 60{t^{3/2}}} \right){\mathbf{i}} - \left( { - 6 + 9{t^{1/2}}} \right){\mathbf{j}} + \left( { - 60{t^4} - 18} \right){\mathbf{k}} \hfill \\
\end{gathered} \]
