#### Answer

\[{f^,}\,\left( x \right) = \frac{{57}}{{\,{{\left( {3x + 10} \right)}^2}}}\]

#### Work Step by Step

\[\begin{gathered}
f\,\left( x \right) = \frac{{6x + 1}}{{3x + 10}} \hfill \\
Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{f^,}\,\left( x \right) \hfill \\
{f^,}\,\left( x \right) = \frac{{\,\left( {3x + 10} \right)\,{{\left( {6x + 1} \right)}^,} - \,\left( {6x + 1} \right)\,{{\left( {3x + 10} \right)}^,}}}{{\,{{\left( {3x + 10} \right)}^2}}} \hfill \\
Then \hfill \\
{f^,}\,\left( x \right) = \frac{{\,\left( {3x + 10} \right)\,\left( 6 \right) - \left( {6x + 1} \right)\,\left( 3 \right)}}{{\,{{\left( {3x + 10} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\
{f^,}\,\left( x \right) = \frac{{18x + 60 - 18x - 3}}{{\,{{\left( {3x + 10} \right)}^2}}} \hfill \\
{f^,}\,\left( x \right) = \frac{{57}}{{\,{{\left( {3x + 10} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]