Answer
$$\frac{{dy}}{{dx}} = \frac{1}{{1-x^2}}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{2}\ln \frac{{1 + x}}{{1 - x}} \cr
& {\text{Use the quotient property for logarithms}} \cr
& y = \frac{1}{2}\left( {\ln \left( {1 + x} \right) - \ln \left( {1 - x} \right)} \right) \cr
& y = \frac{1}{2}\ln \left( {1 + x} \right) - \frac{1}{2}\ln \left( {1 - x} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {1 + x} \right)} \right] - \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {1 - x} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {1 + x} \right)} \right] - \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {1 - x} \right)} \right] \cr
& {\text{Solve the derivatives and simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{1}{{1 + x}}} \right) - \frac{1}{2}\left( {\frac{{ - 1}}{{1 - x}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{1}{{1 + x}} + \frac{1}{{1 - x}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{{1 - x + 1 + x}}{{1-x^2}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1-x^2}} \cr} $$
