Is the module of Kähler differentials a coend?

Let $$\phi\colon R\to S$$ be a ring map. The module of Kähler differentials $$\Omega_{S/R}$$ of $$\phi$$ can be constructed as the following coequaliser: $$\left(\bigoplus_{(a, b)\in S^2} S[(a, b)]\right) \oplus\left( \bigoplus_{(f, g)\in S^2} S[(f, g)]\right)\oplus\left( \bigoplus_{r\in R} S[r] \right)\underset{0}{\rightrightarrows}\bigoplus_{a\in S} S[a],$$ where the above map is defined by \begin{align*} [(a,b)] &\mapsto [a+b]-[a]-[b],\\ [(f,g)] &\mapsto [fg] -f[g]-g[f],\\ [r] &\mapsto [\phi(r)]. \end{align*} This looks strikingly similar to the colimit formula for computing enriched co/ends, which given a $$\mathsf{Mod}_R$$-functor $$P\colon\mathcal{C}^\mathsf{op}\times\mathcal{C}\to\mathsf{Mod}_R$$, takes the form $$\int^{M\in\mathsf{Mod}_R}P(M,M)=\mathrm{Coeq}\left(\bigoplus_{\phi\colon M\to N\in\mathsf{Mod}_R}\mathrm{Hom}_{\mathsf{Mod}_R}(M,N)\otimes_RP(N,M)\rightrightarrows\bigoplus_{M\in\mathsf{Mod}_R}P(M,M)\right).$$

Question. Can we write $$\Omega_{S/R}$$ as a coend in a "nice"* way?

*That is, without directly using the definition of $$\Omega_{S/R}$$ (e.g. using the functor $$\mathrm{Der}_R(S,-)$$ to build the desired coend would certainly be "nice").

Here's a failed short attempt, for what it's worth. Using the Density Theorem and the isomorphism $$\mathrm{Hom}_{\mathsf{Mod}_R}(R,M)\cong M$$, one can write $$\Omega_{S/R}\cong\int^{M\in\mathsf{Mod}_R}(-)\otimes_R\mathrm{Hom}_{\mathsf{Mod}_R}(-,\Omega_{S/R}),$$ but it now seems difficult to give a "nice" description of $$\mathrm{Hom}_{\mathsf{Mod}_R}(-,\Omega_{S/R})$$.