Can we treat $\psi^{c}$ as a field independent from $\psi$? When we derive the Dirac equation from the Lagrangian, $$
\mathcal{L}=\overline{\psi}i\gamma^{\mu}\partial_{\mu}\psi-m\overline{\psi}\psi,
$$
we assume $\psi$ and $\overline{\psi}=\psi^{*^{T}}\gamma^{0}$ are independent. So when we take the derivative of the Lagrangian with respect to $\overline{\psi}$, we get the Dirac equation 
$$
0=\partial_{\mu}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\overline{\psi}\right)}=\frac{\partial\mathcal{L}}{\partial\overline{\psi}}=\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi.
$$
Now if we include a term with charge conjugation, $\psi^{c}=-i\gamma^{2}\psi^{*}$, into the Lagrangian (like $\Delta\mathcal{L}=\overline{\psi}\psi^{c}$), does this $\psi^c$ depend on $\overline{\psi}$ or $\psi$? Why or why not?
If $\psi^{c}$ depends on $\psi$, why wouldn't the reason that $\overline{\psi}$ and $\psi$ are independent apply for $\psi^{c}$ and $\psi$?
If $\psi^{c}$ depends on $\overline{\psi}$, how should we take derivative of $\Delta\mathcal{L}$ with respect to $\overline{\psi}$?
 A: I) The Dirac spinor $\psi$ and its complex conjugate $\psi^*$ are not independent variables, but in some calculations one can treat them as such. 
For the similar question about a complex scalar field $\phi$ and its complex conjugate $\phi^*$, see e.g. this Phys.SE post.
II) The charge conjugated field $\psi^{c}=-i\gamma^{2}\psi^{*}$ is tied to the complex conjugate $\psi^*$ by a bijective transformation, so that they are not independent.
A: Yes, when we want to obtain the equation of motion using Euler-Lagrange equation, we should treat $\psi$ and $\psi^c$ independent, but $\overline{\psi}$ and $\psi^c$ dependent. The reason for this is that we can simply expressed $\psi^c$ in terms of $\overline{\psi}$ by
$$
\psi^{c}=C\overline{\psi}^{T},
$$
where $C=-i\gamma^{2}\gamma^{0}$ is the charge conjugation matrix. So $\overline{\psi}$ and $\psi^c$ are the same degree of freedom.
For the derivative of $\overline{\psi}\psi^{c}$ with respect to $\overline{\psi}$,
one should be really careful because $\psi$ is anticommuting. Since
the derivative in Euler-Lagrange equation actually comes from the
variation of Lagrangian, We should start from the variation
\begin{eqnarray}
\delta\left(\overline{\psi}\psi^{c}\right) & = & \delta\left(\overline{\psi}C\overline{\psi}^{T}\right)=\delta\left(\overline{\psi_{i}}C_{ij}\overline{\psi_{j}}\right)=\delta\left(\overline{\psi_{i}}\right)C_{ij}\overline{\psi_{j}}+\overline{\psi_{i}}C_{ij}\delta\overline{\psi_{j}}\\
 & = & \delta\left(\overline{\psi_{i}}\right)C_{ij}\overline{\psi_{j}}-\delta\left(\overline{\psi_{j}}\right)C_{ij}\overline{\psi_{i}},
\end{eqnarray}
where I use the anticommutation of the fields to get the minus sign
for the last step. Now notice that $C^{T}=C^{+}=-C$. So the last
term is 
\begin{equation}
-\delta\left(\overline{\psi_{j}}\right)C_{ij}\overline{\psi_{i}}=\delta\left(\overline{\psi_{j}}\right)C_{ji}\overline{\psi_{i}}=\delta\left(\overline{\psi_{i}}\right)C_{ij}\overline{\psi_{j}}.
\end{equation}
and we get $\delta\left(\overline{\psi}\psi^{c}\right)=2\delta\left(\overline{\psi}\right)C\overline{\psi}^{T}.$
Therefore, the equation of motion from this term is 
\begin{equation}
\frac{\partial}{\partial\overline{\psi}}\left(\overline{\psi}\psi^{c}\right)=2C\overline{\psi}^{T}=2\psi^{c}.
\end{equation}
