From [nlab](https://ncatlab.org/nlab/show/K%C3%A4hler+differential#AbstractDef), the module of Kähler differentials over some category $\mathcal{C}$ is the free functor:
$$\Omega: \mathcal{C} \to \mathsf{Mod\_{\mathcal{C}}}$$
left-adjoint to the (forgetful) embedding:
$$u: \mathsf{Mod}\_{\mathcal{C}} \cong \mathsf{Ab}(\mathcal{C}) \hookrightarrow \mathcal{C}$$
with $\mathsf{Ab}(\mathcal{C})$ denoting abelian group objects of $\mathcal{C}$, and $\mathsf{Mod}\_{\mathcal{C}}$ denoting the category of modules of $\mathcal{C}$. By this definition, we automatically get a bijection of hom-sets:
$$\mathsf{Mod\_{\mathcal{C}}}(\Omega(R), M) \cong \mathcal{C}(R, u(M))$$
Also, according to the same nlab page as above, $R$-derivations taking values in some $R$-module $M$ (with $R$ some object of $\mathcal{C}$) are morphisms:
$$d: \Omega(R) \to M$$
in $\mathsf{Mod}\_R \cong \mathsf{Mod}\_{R/\mathcal{C}}$. Thus, they could be identified by $\mathcal{C}$-morphisms:
$$X: R \to u(M)$$
because of the adjunction $\Omega \dashv u$. 


At this point, I have two questions:


1. When $\mathcal{C} = \mathsf{CRing}$, the category of commutative and unital rings, do we automatically get the Leibniz/product rule ? Why or why not ?
2. If we do automatically get the Leibniz rule, then is it also the case in categories more general than $\mathsf{CRing}$ ?


Thank you.



Categorical Kähler differentials and the Leibniz rule
1. The Leibniz rule follows immediately from the last description
of derivations as morphisms of commutative rings X:R→u(M).


Indeed, u(M) is the square-zero extension of some R-module M'
(in the traditional sense), i.e., u(M)=R⊕M'.


Now a morphism of commutative rings f:R→R⊕M' in the slice category C/R
(not in C, as is claimed in the main post) necessarily has the form
r↦(r,φ(r)), for some linear map φ.
Since f is a homomorphism, we have f(1)=(1,φ(1))=(1,0), so φ(1)=0.
Also $$(rr',φ(rm'+r'm))=f(rr',rm'+r'm)=f((r,m)(r',m'))=f(r,m)f(r',m')=(r,φ(m))(r',φ(m'))=(rr',rφ(m')+r'φ(m)),$$
so $$φ(rm'+r'm)=rφ(m')+r'φ(m).$$


This is precisely the Leibniz rule.


2. Yes, for instance, this is true for algebras over Fermat theories.
See Carchedi and Roytenberg's [Homological Algebra for Superalgebras of Differentiable Functions](https://arxiv.org/abs/1212.3745).


