Answer
\[{f^,}\,\left( x \right) = \frac{1}{x} - \frac{{6x}}{{{x^2} + 1}}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = \ln \frac{{2x}}{{\,{{\left( {{x^2} + 1} \right)}^3}}} \hfill \\
\hfill \\
Use\,\,\,the\,\,property\,\,\ln \,\left( {\frac{a}{b}} \right) = \ln a - \ln b \hfill \\
\hfill \\
then \hfill \\
\hfill \\
f\,\left( x \right) = \ln 2x - \ln \,{\left( {{x^2} + 1} \right)^3} \hfill \\
\hfill \\
f\,\left( x \right) = \ln 2x - 3\ln \,\left( {{x^2} + 1} \right) \hfill \\
\hfill \\
Differentiate{\text{ }}Use\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{2}{{2x}} - 3\,\left( {\frac{{2x}}{{{x^2} + 1}}} \right) \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{1}{x} - \frac{{6x}}{{{x^2} + 1}} \hfill \\
\end{gathered} \]
