http://en.wikipedia.org/wiki/Quantum_electrodynamics#Equations_of_motion

 

When you take the variational derivative, of the QED Lagrangian, with respect to the wave function [math]\Psi[/math]...

 

why doesn't the derivative include terms, due to the conjugate transpose of the wave function [math]\bar{\Psi}[/math] ?

 

In Classical analogy, for a Lagrangian with the KE term [math]\left( \frac{1}{2m} \vec{p}^T \circ \vec{p} \right)[/math], derivatives with respect to momentum would include (one) terms, from the transpose of momentum, which is essentially the same mathematical object

You can treat them as independent variables, and you should be thinking in terms of fermionic fields here not wavefunctions.