Going from the Dirac Lagrangian to the adjoint Dirac equation I am comfortable doing the following calculation, Derivation of the adjoint of Dirac equation, notably — going from the standard Dirac equation to the adjoint Dirac equation via using Dirac conjugation.
I am however not comfortable deriving the adjoint Dirac equation from the Euler Lagrange equation (for $\psi$, the EL eq for $\overline{\psi}$ leads to the standard Dirac equation) of the Dirac Lagrangian 
$$\mathcal{L} =\overline{\psi}(i\gamma^\mu \partial_\mu -m)\psi $$.
My problem boils down to the following term, 
$$
\partial_\mu (i\overline{\psi}\gamma^\mu)$$
How can I get this to yield 
$$
-i\gamma^\mu \partial_\mu \overline{\psi}?
$$
My only thought is to write out Dirac conjugate, write out the Einstein summation, and use the properties of the gamma matrices ($\gamma^0\gamma^0=$ the identity matrix, and $\gamma^0\gamma^i=-\gamma^i\gamma^0$), but it doesn't seem to yield anything fruitful.
Thoughts?
Cheers
 A: The Euler-Lagrange-Equation is given by:
$$\frac{\partial\mathcal{L}}{\partial {\psi}} - \partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi)} = 0$$
Let us take both derivatives separately. We treat $\psi$ and $\bar{\psi}$ as independent fields. This gives
$$\frac{\partial}{\partial\psi}\bar{\psi}(i\gamma^\mu\partial_{\mu}-m)\psi = -m \bar{\psi}$$
and
$$\partial_{\mu}\frac{\partial}{\partial(\partial_{\mu}\psi)}\bar{\psi}(i\gamma^\mu\partial_{\mu}-m)\psi = i\partial_{\mu}\bar{\psi}\gamma^\mu$$
So by plugging this into the first equation we get:
$$-m\bar{\psi} - i\partial_{\mu}\bar{\psi}\gamma^\mu = 0 $$
Now, to make this look nicer (and bring this in the usual form) we say, that the differential operator operates to the left:
$$\bar{\psi}(i\gamma^{\mu} \overleftarrow{\partial}_{\mu}+m) = 0$$
Your proposed solution $i \partial_\mu \gamma^\mu \bar{\psi}$ doesn't work dimensional wise. Since $\bar{\psi}$ is a row vector, it needs to be left of the $\gamma^\mu$, which is a $4\times 4$ square matrix.
