Answer
$$y' = \frac{1}{x}$$
Work Step by Step
$$\eqalign{
& y = \ln 2x \cr
& {\text{find the derivative}} \cr
& y' = \left( {\ln 2x} \right)' \cr
& {\text{use }}\left( {\ln u} \right)' = \frac{{u'}}{u},{\text{ }}\operatorname{in} {\text{ this exercise }}u = 2x \cr
& y' = \frac{{\left( {2x} \right)'}}{{2x}} \cr
& {\text{compute derivative}} \cr
& y' = \frac{2}{{2x}} \cr
& {\text{simplifying}} \cr
& y' = \frac{1}{x} \cr} $$
