Answer
\[{f^,}\,\left( x \right) = \frac{{12}}{{3x + 1}}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = \ln \,{\left( {3x + 1} \right)^4} \hfill \\
\hfill \\
Use\,\,\ln {a^u} = u\ln a \hfill \\
\hfill \\
f\,\left( x \right) = 4\ln \,\left( {3x + 1} \right) \hfill \\
\hfill \\
Differentiate{\text{ }}Use\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
{f^,}\,\left( x \right) = 4\,\left( {\frac{3}{{3x + 1}}} \right) \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{12}}{{3x + 1}} \hfill \\
\end{gathered} \]
