#### Answer

\[ = \frac{{2x}}{{{x^2} - 1}}\]

#### Work Step by Step

\[\begin{gathered}
\frac{d}{{dx}}\,\left( {\ln \,\left| {{x^2} - 1} \right|} \right) \hfill \\
\hfill \\
Use\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
set\,\,u = {x^2} - 1 \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
= \frac{{2x}}{{{x^2} - 1}} \hfill \\
\end{gathered} \]