How to calculate Noether current for Yang Mills theory The Lagrangian of the Yang-Mills fields is given by 
\begin{align}
\mathcal{L}=-\frac{1}{4} F^a_{\mu\nu}~ F_a^{\mu\nu} +\bar{\psi}(i\gamma^{\mu} 
  D_{\mu})\psi- m~ \psi \bar{\psi}
\end{align}
where: 
\begin{align}
D_\mu \psi = \partial_\mu \psi - i g~ t^a_{ij}~ A_\mu^a~ \psi\\
F_{\mu\nu}^a = \partial_\mu A^a_\nu - \partial_\nu A_\mu^a + i g f^{abc} A^b_\mu A_\nu^c
\end{align}
I try to get the Noether current here: so starting by the equations of motions: 
$ \frac{\partial \mathcal{L}}{\partial\psi} - \partial_\mu  \frac{\partial \mathcal{L}}{\partial(\partial_\mu\psi)} = 0  ~~~~~~~~~~(1)\\  \frac{\partial \mathcal{L}}{\partial\bar{\psi}} - \partial_\mu  \frac{\partial \mathcal{L}}{\partial(\partial_\mu\bar{\psi})} = 0~~~~~~~~~(2) \\  \frac{\partial \mathcal{L}}{\partial A_\mu} - \partial_\nu  \frac{\partial \mathcal{L}}{\partial(\partial_\nu A_\mu)} = 0 ~~~~~~~(3)\\ $ 
These yield:
$ g \bar{\psi} \gamma^\mu A_\mu^a - m \bar{\psi}  -i \partial_\mu \bar{\psi} \gamma^\mu = 0~~~~~~~~(4)\\
g \gamma^\mu A_\mu^a \psi - m \psi  = 0~~~~~~~~~~~(5)\\
g\bar{\psi} \gamma^\mu t^a_{ij} \psi_j + g f^{abc} A^b_\nu F^{c~\mu\nu}= \partial_\nu F^{a~ \nu \mu}~~~~~~~~(6) $
Clearly (6) is the right equation of motion of the Yang Mills theory with a conserved current : $J^{a~ \mu} = g\bar{\psi} \gamma^\mu t^a_{ij} \psi_j$ , see for instance Peskin's book Equation(15.51) . Now I have extra terms in (4) and (5) what's wrong I made?
 A: The current is just the LHS of your eq. $(6)$. In fact, the antisimmetry of $F^{\mu \nu}$ implies  $\partial _\mu \partial _\nu F^{\mu \nu}=0$.
Observe that the gauge fields carry a non-zero charge, since they transform non-trivially under global transformations. In fact, they transform according to the adjoint representation of the group of simmetry. The part of the current which involves only $\psi$ is the covariant current $\mathscr J ^\mu$, which enters in:$$D_\mu F^{\mu \nu} =\mathscr J ^\nu.$$ This current (differently from the Noether current $J^\mu$) transforms like the Yang-Mills tensor under gauge transformations. 
A: You have forgotten to add the term
$i\partial_\mu\psi \gamma^\mu$
in (5).
Equation (4) is correct. Note that $D_\mu = \partial_\mu + iA_\mu$.
A: In the Lagrangian of the question, there is a typo, it should be $\bar{\psi}\psi$ for the last term.
The equation (4) is basically correct except the sign of $i \gamma^{\mu} \partial_{\mu} \psi$ should be positive. By taking complex conjugate of it , you get the equation of motion as $i \gamma^{\mu} D_{\mu} \psi - m \psi =0$.
The equation (5) should be the same when you do a partial integral for the term $\bar{\psi} i \gamma^{\mu} \partial_{\mu} \psi=i\gamma^{\mu} \partial{\left(\bar{\psi}\psi\right)}- \partial{\bar{\psi} i \gamma^{\mu} \psi}$ and then compute the Euler-Lagrange equation.
The equation (6) is basically correct. It should be $\partial_{\nu} F^{a \nu \mu} + g \bar{\psi} \gamma^{\mu} t^a \psi + i g F^{c \mu \nu} f^{abc} A^b_{\nu}=0$.
There are two equations of motion in total for two fields, which makes sense.
