Let $A$ be an algebra over a ring $R$. Let $\mathfrak P$ be a prime of $A$ with preimage $\mathfrak p$ in $R$. We have the localized ring $A_{\mathfrak p}$, which further localizes to $A_{\mathfrak P}$.
Let $\Omega_{A/R}$ be the module of differentials of $A$ over $R$. There are two very nice properties of $\Omega_{A/R}$. First, there is base change: if $R'$ is an $R$-algebra, then
$$\Omega_{A/R} \otimes_R R' = \Omega_{(A\otimes_R R')/R'} \tag{1}$$
Second, there is localization: if $S \subset A$ is multiplicatively closed, then
$$S^{-1}\Omega_{A/R} = \Omega_{S^{-1}A/R} \tag{2}$$ or in other words, $\Omega_{A/R} \otimes_A S^{-1}A = \Omega_{S^{-1}A/R}$.
I want to combine these properties to say that $(\Omega_{A/R})_{\mathfrak P} = \Omega_{A_{\mathfrak P}/R_{\mathfrak p}}$ as $A_{\mathfrak P}$-modules. I haven't seen this property written down before, but I believe it is true. On account of the transitivity of localization, we have
$$(\Omega_{A/R})_{\mathfrak P} = ((\Omega_{A/R})_{\mathfrak p})_{\mathfrak P} = (\Omega_{A/R} \otimes_R R_{\mathfrak p})_{\mathfrak P} \stackrel{(1)}{=} (\Omega_{A \otimes_R R_{\mathfrak p}/R_{\mathfrak p}})_{\mathfrak P} = (\Omega_{A_{\mathfrak p}/R_{\mathfrak p}})_{\mathfrak P} \stackrel{(2)}{=} \Omega_{(A_{\mathfrak p})_{\mathfrak P}/R_{\mathfrak p}} = \Omega_{A_{\mathfrak P}/R_{\mathfrak p}}$$
Is my reasoning correct here?