Answer
$$\frac{{dy}}{{dx}} = \frac{{{e^{2x}}}}{{1 + {e^{4x}}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{2}\arctan {e^{2x}} \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{1}{2}\arctan {e^{2x}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\frac{d}{{dx}}\left[ {\arctan {e^{2x}}} \right] \cr
& {\text{Use the rule }}\frac{d}{{dx}}\left[ {\arctan u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{1}{{1 + {{\left( {{e^{2x}}} \right)}^2}}}} \right)\frac{d}{{dx}}\left[ {{e^{2x}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{1}{{1 + {{\left( {{e^{2x}}} \right)}^2}}}} \right)\left( {2{e^{2x}}} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{{e^{2x}}}}{{1 + {e^{4x}}}} \cr} $$
