#### Answer

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

#### Work Step by Step

\[\begin{gathered}
\frac{d}{{dx}}\,\,\left[ {\ln \,\left( {\frac{{x + 1}}{{x - 1}}} \right)} \right] \hfill \\
\hfill \\
Use\,\,the\,\,\,property\,\,\ln \,\left( {\frac{a}{b}} \right) = \ln a - \ln b \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
\frac{d}{{dx}} = \,\,\left[ {\ln \,\left( {x + 1} \right) - \ln \,\left( {x - 1} \right)} \right] \hfill \\
\hfill \\
Differentiate\,use\,\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
= \frac{1}{{x + 1}} - \frac{1}{{x - 1}} \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
= \frac{{x - 1 - x - 1}}{{{x^2} - 1}} \hfill \\
\hfill \\
so \hfill \\
\hfill \\
= - \frac{2}{{{x^2} - 1}} \hfill \\
\hfill \\
R = \left\{ { - 1,1} \right\} \hfill \\
\hfill \\
\end{gathered} \]