#### Answer

\[\,\left( { - \infty ,0} \right) \cup \,\left( {0,\infty } \right)\]

#### Work Step by Step

\[\begin{gathered}
\frac{d}{{dx}}\,\left( {\ln 2{x^8}} \right) \hfill \\
\hfill \\
Use\,\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{d}{{dx}}\,\left( {\ln 2{x^8}} \right) = \frac{{\frac{d}{{dx}}\left[ {\ln 2{x^8}} \right]}}{{2{x^8}}} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \frac{{16{x^7}}}{{2{x^8}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{8}{{{x}}} \hfill \\
\hfill \\
Interval \hfill \\
\hfill \\
\,\left( { - \infty ,0} \right) \cup \,\left( {0,\infty } \right) \hfill \\
\end{gathered} \]