Answer
\[{y^,} =- \frac{{15x}}{{3x + 2}} - 5\ln \,\left( {3x + 2} \right)\]
Work Step by Step
\[\begin{gathered}
y = - 5x\ln \,\left( {3x + 2} \right) \hfill \\
Differentiate \hfill \\
{y^,} = \,\,\left[ { - 5x\ln \,\left( {3x + 2} \right)} \right] \hfill \\
Use\,\,the\,\,product\,\,rule \hfill \\
{y^,} = - 5x\,\,{\left[ {\ln \,\left( {3x + 2} \right)} \right]^,} + \ln \,\left( {3x + 2} \right)\,{\left( { - 5x} \right)^,} \hfill \\
Use\,\,the\,\,formula \hfill \\
\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\
Then \hfill \\
{y^,} = - 5x\,\left( {\frac{{\,{{\left( {3x + 2} \right)}^,}}}{{3x + 2}}} \right) + \ln \,\left( {3x + 2} \right)\,\left( { - 5} \right) \hfill \\
{y^,} = - 5x\,\left( {\frac{3}{{3x + 2}}} \right) - 5\ln \,\left( {3x + 2} \right) \hfill \\
Multiply \hfill \\
{y^,} = -\frac{{15x}}{{3x + 2}} - 5\ln \,\left( {3x + 2} \right) \hfill \\
\end{gathered} \]
