D'Alembert Operator on Fermionic Field in Path Integral

I am learning the Faddeev–Popov path integral formlism with Schwartz's QFT textbook. In the section 25.4.2 "BRST invariance", I came across the Lagrangian as: $$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^{2}+\left(D_{\mu} \phi_{i}^{\star}\right)\left(D_{\mu} \phi_{i}\right)-m^{2} \phi_{i}^{\star} \phi_{i}-\frac{1}{2 \xi}\left(\partial_{\mu} A_{\mu}\right)^{2}-\bar{c} \square c$$ And it says the equation of motion of the fermionic fields satisify: $$\square c=\square \bar{c}=0$$ I am curious and surprised with the derivative of the fermionc field(which certainly is a grassmann-valued function with respect to space-time coordinates.) But the whole Schwartz's textbook doesn't mention how to define the derivative of the grassmann-valued function with respect to certain real-valued variables(There is indeed a section about Grassmann Algebra, but it doesn't introduce how to define this!).

So my question is it correct to define the derivative as similar to ordinary calculus: $$c'(x)=\lim_{\epsilon \to 0}\frac{c(x+\epsilon)-c(x)}{\epsilon}$$ I suspect that the above "definition" is not well-defined. Could anyone give a physical-intution explanation to the derivative, or any reference to the strange derivative?