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