Let $R \to A$ be a morphism of rings, making $A$ into an $R$-algebra. Denote the module of differentials of $A$ over $R$ by $\Omega_{A/R}$, with exterior derivative $d_{A/R}: A \to \Omega_{A/R}$. Let $R \to R'$ be a ring homomorphism and denote $A' = A \otimes_R R'$. I am trying to show that $\Omega_{A'/R'} = \Omega_{A/R} \otimes_A A'$, with exterior derivative $d_{A'/R'} = d_{A/R} \otimes_R id_{R'}$ - here, I am using the identification $\Omega_{A/R} \otimes_R R' \simeq \Omega_{A/R} \otimes_A A'$ with $m \otimes_R r' = m \otimes_A (1 \otimes_R r')$, for $m \in \Omega_{A/R}, r' \in R'$. Thus, $d_{A'/R'}(x \otimes r') = d_{A/R}(x) \otimes_A (1 \otimes_R r')$, for $x \in A, r' \in R'$. I then tried to show that $(\Omega_{A/R} \otimes_A A', d_{A/R} \otimes_R id_{R'})$ satisfies the universal property defining the module of differentials. Here's what I've done so far:
Let $M$ be an $A'$-module and $d: A' \to M$ be a derivation of $A'$ into $M$ over $R'$. Then, considering the canonical morphism $\iota: A \to A', x \mapsto x \otimes1$, we can make $M$ into an $A$-module and $\tilde{d} = d \circ \iota$ is a derivation of $A$ into $M$ over $R$. By the universal property of the module of differentials, there is a homomorphism of $A$-modules $\varphi: \Omega_{A/R} \to M$ such that $\varphi \circ d_{A/R} = \tilde{d}$. I tried to argue that $\varphi \otimes_A id_{A'}$ would be the unique morphism $\Omega_{A/R} \otimes_A A' \to M$ satisfying the universal property - but the target of $\varphi \otimes_A id_{A'}$ is $M \otimes_A A'$, not $M$. If there was a canonical homomorphism of $A'$-modules $M \otimes_A A' \to M$, I could maybe use it to finish the proof, but I don't know if such morphism exists. Maybe I am going the wrong way.
Edit: Multiplication by scalar gives an obvious morphism $M \otimes_A A' \to M$, so I used this composed with $\varphi \otimes_A id_{A'}$ to define a map $\psi$ and showed that $\psi \circ d_{A'/R'} = d$. It remains to show uniqueness of $\psi$, I would appreciate some help.
