Answer
$$\frac{1}{{{x^2} - 1}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\ln \sqrt {\frac{{x - 1}}{{x + 1}}} } \right] \cr
& {\text{radical properties}} \cr
& \frac{d}{{dx}}\left[ {\ln \frac{{\sqrt {x - 1} }}{{\sqrt {x + 1} }}} \right] \cr
& \frac{d}{{dx}}\left[ {\ln \frac{{{{\left( {x - 1} \right)}^{1/2}}}}{{{{\left( {x + 1} \right)}^{1/2}}}}} \right] \cr
& {\text{logarithm of a quotient}} \cr
& = \frac{d}{{dx}}\left[ {\ln {{\left( {x - 1} \right)}^{1/2}} - \ln {{\left( {x + 1} \right)}^{1/2}}} \right] \cr
& {\text{power rule for logarithms}} \cr
& = \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {x - 1} \right) - \frac{1}{2}\ln \left( {x + 1} \right)} \right] \cr
& {\text{differentiate}} \cr
& \frac{1}{2}\left( {\frac{1}{{x - 1}}} \right) - \frac{1}{2}\left( {\frac{1}{{x + 1}}} \right) \cr
& {\text{simplify}} \cr
& = \frac{1}{2}\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) \cr
& = \frac{1}{2}\left( {\frac{{x + 1 - x + 1}}{{{x^2} - 1}}} \right) \cr
& = \frac{1}{2}\left( {\frac{2}{{{x^2} - 1}}} \right) \cr
& = \frac{1}{{{x^2} - 1}} \cr} $$