# Does d/d(spinor) anticommute with a spinor?

Weyl spinors anticommute (see, e.g. Why isn't the anticommutativity of spinors sufficient as "spin-statistics-theorem"?).

If we consider the derivative, with respect to a Weyl spinor $\frac{d}{d\chi}$, does this derivative anticommute with a given Weyl spinor $\chi$?

To be explicit, consider the product

$$\frac{d}{d\chi} \chi f(x) \underbrace{=}_{\text{product rule}} \left(\frac{d}{d\chi} \chi \right ) f(x) \pm \chi \left(\frac{d}{d\chi} f(x) \right ) ,$$ where $f(x)$ is an arbitrary object, such that $\frac{d}{d\chi} f(x) \neq 0$. Which sign here is correct? Does the product rule for spinors involve a minus sign instead of a plus sign?

2. Secondly, a Grassmann-odd derivative $\frac{\partial^L}{\partial \chi}$ does not always anticommute with a Grassmann-odd variable $\tilde{\chi}$, cf. the title question (v1). E.g. if $\tilde{\chi}$ depends on $\chi$.
3. More generally, the graded Leibniz rule for a derivative acting from left ($L$) reads $$\frac{\partial^L(fg)}{\partial \chi}~=~\frac{\partial^Lf}{\partial \chi}g+(-1)^{|\chi||f|}f\frac{\partial^Lg}{\partial \chi},$$ while for a derivative acting from right ($R$) it is $$\frac{\partial^R(fg)}{\partial \chi}~=~f\frac{\partial^Rg}{\partial \chi}+(-1)^{|\chi||g|}\frac{\partial^Rf}{\partial \chi}g.$$ Here $|f|$ denotes the Grassmann parity of $f$, and $\chi$ is any type of variable (not necessarily a spinor). See also this related Phys.SE post.
4. For completeness, let us mention that the left & right derivatives are equal up to a sign factor: $$\frac{\partial^Lf}{\partial\chi}~=~(-1)^{(|f|+1)|\chi|}\frac{\partial^Rf}{\partial\chi}.$$
Actually there are two different definitions of $d\over d\psi$ for spinor $\psi$, whether acting from right or left. It satisfies the calculus of Grasmann algebra(anticommuting numbers). If you take the definition of $d\over d\psi$ to act from left, then if $f(x)$ is commutative number(c-number in literature usually), then it should be \begin{equation} \frac{d}{d\chi} \chi f(x) \underbrace{=}_{\text{product rule}} \left(\frac{d}{d\chi} \chi \right ) f(x) + \chi \left(\frac{d}{d\chi} f(x) \right ) \end{equation} or if $f(x)$ contains odd number of spinors, then it's a Grassmann number, the sign should be minus.