Answer
$\frac{{dy}}{{dx}} = \frac{2}{{1 - 4{x^2}}}$
Work Step by Step
$$\eqalign{
& y = \ln \sqrt {\frac{{1 + 2x}}{{1 - 2x}}} \cr
& {\text{Use the radical property }}\sqrt m = {m^{1/2}} \cr
& y = \ln {\left( {\frac{{1 + 2x}}{{1 - 2x}}} \right)^{1/2}} \cr
& {\text{Use the logarithmic property }}\ln {a^n} = n\ln a \cr
& y = \frac{1}{2}\ln \left( {\frac{{1 + 2x}}{{1 - 2x}}} \right) \cr
& {\text{Use the logarithmic property }}\ln \left( {\frac{a}{b}} \right) = \ln a - \ln b \cr
& y = \frac{1}{2}\ln \left( {1 + 2x} \right) - \frac{1}{2}\ln \left( {1 - 2x} \right) \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {1 + 2x} \right)} \right] - \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {1 - 2x} \right)} \right] \cr
& {\text{Pull out the constants}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {1 + 2x} \right)} \right] - \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {1 - 2x} \right)} \right] \cr
& {\text{Compute derivatives}}{\text{, use }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{1}{{1 + 2x}}} \right)\left( 2 \right) - \frac{1}{2}\left( {\frac{1}{{1 - 2x}}} \right)\left( { - 2} \right) \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + 2x}} + \frac{1}{{1 - 2x}} \cr
& \frac{{dy}}{{dx}} = \frac{{1 - 2x + 1 + 2x}}{{\left( {1 + 2x} \right)\left( {1 - 2x} \right)}} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{1 - 4{x^2}}} \cr} $$
