#### Answer

\[{f^,}\,\left( t \right) = \,\left( {6t{e^{{t^2}}} + 15} \right)\,{\left( {{e^{{t^2}}} + 5t} \right)^2}\]

#### Work Step by Step

\[\begin{gathered}
f\,\left( t \right) = \,{\left( {{e^{{t^2}}} + 5t} \right)^3} \hfill \\
Use\,\,the\,\,generalized\,\,power\,\,rule \hfill \\
\frac{d}{{dx}}\,\,{\left[ {g\,\left( x \right)} \right]^n} = n\,\,{\left[ {g\,\left( x \right)} \right]^{n - 1}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{f^,}\,\left( t \right) = 3{\left( {{e^{{t^2}}} + 5t} \right)^{3 - 1}}\,{\left( {{e^{{t^2}}} + 5t} \right)^,} \hfill \\
Differentiate\,\,{e^{{t^2}}} + 5t \hfill \\
{f^,}\,\left( t \right) = 3{\left( {{e^{{t^2}}} + 5t} \right)^2}\,\left( {2t{e^{{t^2}}} + 5} \right) \hfill \\
Multiplying \hfill \\
{f^,}\,\left( t \right) = \,\left( {6t{e^{{t^2}}} + 15} \right)\,{\left( {{e^{{t^2}}} + 5t} \right)^2} \hfill \\
\end{gathered} \]