Answer
\[{y^,} = \frac{{2 - 24{x^2}}}{{2x - 8{x^3}}}\]
Work Step by Step
\[\begin{gathered}
y = \ln \left| { - 8{x^3} + 2x} \right| \hfill \\
Find\,\,the\,\,derivative \hfill \\
{y^,} = \,\,\left[ {\ln \,\left| { - 8{x^3} + 2x} \right|} \right] \hfill \\
Use\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\
Here\,\,g\,\left( x \right) = - 8{x^3} + 2x \hfill \\
Then \hfill \\
{y^,} = \frac{{\,{{\left( { - 8{x^3} + 2x} \right)}^,}}}{{ - 8{x^3} + 2x}} \hfill \\
Differentiating \hfill \\
{y^,} = \frac{{ - 24{x^2} + 2}}{{ - 8{x^3} + 2x}} \hfill \\
{y^,} = \frac{{2 - 24{x^2}}}{{2x - 8{x^3}}} \hfill \\
\end{gathered} \]