Answer
$$y' = 2\operatorname{sech} 2x{\text{csch}}2x$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\tanh 2x} \right) \cr
& {\text{find the derivative}} \cr
& y' = \left[ {\ln \left( {\tanh 2x} \right)} \right]' \cr
& {\text{use }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{{u'}}{u} \cr
& y' = \frac{{\left( {\tanh 2x} \right)'}}{{\tanh 2x}} \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{ see page 476}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {\tanh u} \right] = {\operatorname{sech} ^2}u\frac{{du}}{{dx}} \cr
& y' = \frac{{{{\operatorname{sech} }^2}2x\left( 2 \right)}}{{\tanh 2x}} \cr
& {\text{simplifying}} \cr
& y' = 2\left( {\frac{1}{{{{\cosh }^2}2x}}} \right)\left( {\frac{{\cosh 2x}}{{\sinh 2x}}} \right) \cr
& y' = 2\left( {\frac{1}{{\cosh 2x}}} \right)\left( {\frac{1}{{\sinh 2x}}} \right) \cr
& y' = 2\operatorname{sech} 2x{\text{csch}}2x \cr} $$
