#### Answer

\[{y^,} = \frac{{3{x^2}}}{{1 + {x^3}}}\]

#### Work Step by Step

\[\begin{gathered}
y = \ln \,\left( {1 + {x^3}} \right) \hfill \\
Find\,\,the\,\,derivative \hfill \\
{y^,} = \,\,\left[ {\ln \,\left( {1 + {x^3}} \right)} \right] \hfill \\
Use\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\
Here\,\,g\,\left( x \right) = 1 + {x^3}\, \hfill \\
Then \hfill \\
{y^,} = \frac{{\,{{\left( {1 + {x^3}} \right)}^,}}}{{1 + {x^3}}} \hfill \\
Differentiating \hfill \\
{y^,} = \frac{{3{x^2}}}{{1 + {x^3}}} \hfill \\
\end{gathered} \]