#### Answer

\[{v^,} = \frac{{1 - 3\ln u}}{{{u^4}}}\]

#### Work Step by Step

\[\begin{gathered}
V = \frac{{\ln u}}{{{u^3}}} \hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
{v^,} = \frac{{{u^3}\,\,{{\left[ {\ln u} \right]}^,} - \,\,\left[ {\ln u} \right]\,\,{{\left[ {{u^3}} \right]}^,}}}{{\,\,{{\left[ {{u^3}} \right]}^2}}} \hfill \\
Where \hfill \\
\ln u = \frac{{{u^,}}}{u} = \frac{1}{u} \hfill \\
Then \hfill \\
{v^,} = \frac{{{u^3}\,\left( {\frac{1}{u}} \right) - \ln u\,\left( {3{u^2}} \right)}}{{{u^6}}} \hfill \\
Simplify \hfill \\
{v^,} = \frac{{{u^2} - 3{u^2}\ln u}}{{{u^6}}} = \frac{{1 - 3\ln u}}{{{u^4}}} \hfill \\
\hfill \\
\hfill \\
\end{gathered} \]