#### Answer

\[{f^,}\,\left( t \right) = \frac{{2t}}{{\,{{\left( {{t^2} + 3} \right)}^2}}}\]

#### Work Step by Step

\[\begin{gathered}
f\,\left( t \right) = \frac{{4{t^2} + 11}}{{{t^2} + 3}} \hfill \\
Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{f^,}\,\left( t \right) \hfill \\
{f^,}\,\left( t \right) = \frac{{\,\left( {{t^2} + 3} \right)\,{{\left( {4{t^2} + 11} \right)}^,} - \,\left( {4{t^2} + 11} \right)\,{{\left( {{t^2} + 3} \right)}^,}}}{{\,{{\left( {{t^2} + 3} \right)}^2}}} \hfill \\
Then, \hfill \\
{f^,}\,\left( t \right) = \frac{{\,\left( {{t^2} + 3} \right)\,\left( {8t} \right) - \left( {4{t^2} + 11} \right)\,\left( {2t} \right)}}{{\,{{\left( {{t^2} + 3} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\
{f^,}\,\left( t \right) = \frac{{8{t^3} + 24t - 8{t^3} - 22t}}{{\,{{\left( {{t^2} + 3} \right)}^2}}} \hfill \\
{f^,}\,\left( t \right) = \frac{{2t}}{{\,{{\left( {{t^2} + 3} \right)}^2}}} \hfill \\
\end{gathered} \]