#### Answer

\[\frac{{dy}}{{dx}} = \frac{{3\pi {x^2}}}{{{x^3} + 1}}\]

#### Work Step by Step

\[\begin{gathered}
y = \ln \,{\left( {{x^3} + 1} \right)^\pi } \hfill \\
\hfill \\
Use\,\,\,the\,\,property\,\,\ln {a^u} = u\ln a \hfill \\
\hfill \\
y = \pi \ln \,\left( {{x^3} + 1} \right) \hfill \\
\hfill \\
Use\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
y = \pi \frac{{{{\left( {{x^3} + 1} \right)}^,}\,}}{{{x^3} + 1}} \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \pi \,\left( {\frac{{3{x^2}}}{{{x^3} + 1}}} \right) \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{3\pi {x^2}}}{{{x^3} + 1}} \hfill \\
\end{gathered} \]