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