Answer
$H'\left( v \right) = 2{e^{\tanh 2v}}{\operatorname{sech} ^2}2v$
Work Step by Step
$$\eqalign{
& H\left( v \right) = {e^{\tanh 2v}} \cr
& {\text{Differentiate}} \cr
& H'\left( v \right) = \frac{d}{{dv}}\left[ {{e^{\tanh 2v}}} \right] \cr
& {\text{Use }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}} \cr
& H'\left( v \right) = {e^{\tanh 2v}}\frac{d}{{dv}}\left[ {\tanh 2v} \right] \cr
& {\text{Use the Derivatives of Hyperbolic Functions }} \cr
& \frac{d}{{dv}}\left[ {\tanh u} \right] = {\operatorname{sech} ^2}u\frac{{du}}{{dv}},{\text{ let }}u = 2v,{\text{ so}} \cr
& H'\left( v \right) = {e^{\tanh 2v}}\left[ {{{\operatorname{sech} }^2}2v\frac{d}{{dv}}\left[ {2v} \right]} \right] \cr
& {\text{Compute the derivative and simplify}} \cr
& H'\left( v \right) = {e^{\tanh 2v}}\left[ {{{\operatorname{sech} }^2}2v\left( 2 \right)} \right] \cr
& H'\left( v \right) = 2{e^{\tanh 2v}}{\operatorname{sech} ^2}2v \cr} $$
