Answer
\[{y^,}\, = \frac{{ - 17}}{{\,{{\left( {4 + t} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{5 - 3t}}{{4 + t}} \hfill \\
Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{y^,}\, \hfill \\
{y^,} = \frac{{\,\left( {4 + t} \right)\,{{\left( {5 - 3t} \right)}^,} - \,\left( {5 - 3t} \right)\,{{\left( {4 + t} \right)}^,}}}{{\,{{\left( {4 + t} \right)}^2}}} \hfill \\
Then \hfill \\
{y^,}\, = \frac{{\left( {4 + t} \right)\,\left( { - 3} \right) - \,\left( {5 - 3t} \right)\,\left( 1 \right)}}{{\,{{\left( {4 + t} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\
{y^,}\, = \frac{{ - 12 - 3t - 5 + 3t}}{{\,{{\left( {4 + t} \right)}^2}}} \hfill \\
{y^,}\, = \frac{{ - 17}}{{\,{{\left( {4 + t} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]