Answer
\[{y^,} = \frac{{\frac{{ - 6x + 2}}{x} + 6\ln x}}{{\,{{\left( {3x - 1} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{ - 2\ln x}}{{3x - 1}} \hfill \\
Differentiate \hfill \\
{y^,} = \,\,\left[ {\frac{{ - 2\ln x}}{{3x - 1}}} \right] \hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
{y^,} = \frac{{\,\left( {3x - 1} \right)\,{{\left( { - 2\ln x} \right)}^,}\,\left( { - 2\ln x} \right)\,{{\left( {3x - 1} \right)}^,}}}{{\,{{\left( {3x - 1} \right)}^2}}} \hfill \\
\operatorname{Re} call\,\,that\,\,\frac{d}{{dx}}\,\,\left[ {\ln x} \right] = \frac{1}{x} \hfill \\
{y^,} = \frac{{\,\left( {3x - 1} \right)\,\left( { - 2} \right)\,\left( {\frac{1}{x}} \right) + 2\ln x\,\left( 3 \right)}}{{\,{{\left( {3x - 1} \right)}^2}}} \hfill \\
Multiplying \hfill \\
{y^,} = \frac{{\frac{{ - 6x + 2}}{x} + 6\ln x}}{{\,{{\left( {3x - 1} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]
