Proof that $\Omega_{S^{-1}B/A}=S^{-1}\Omega_{B/A}$ The notations are: $\varphi:A\to B$ is a ring map, $S\subset B$ is a multiplicative subset of $B$, and $\Omega_{B/A}$ is the module of Kähler $A$-differentials of $B$.
In a proof of the fact that $\Omega_{S^{-1}B/A}=S^{-1}\Omega_{B/A}$, the document I am reading says that there is a canonical map from $S^{-1}\Omega_{B/A}$ to $\Omega_{S^{-1}B/A}$, and that to show it is an isomorphism, it suffices to construct an $A$-derivation from $S^{-1}B$ to $S^{-1}\Omega_{B/A}$.
I know that this will yield a map $\Omega_{S^{-1}B/A}\to S^{-1}\Omega_{B/A}$, but I can't see why this would be an inverse of the canonical map. I guess that this is just a universal property argument, but I can't figure out how to write things down.
Any help is appreciated.
 A: $\Omega_{B/A}$ is a $B$-module so $S^{-1}\Omega_{B/A}$ is an $S^{-1}B$-module. Let $U$ denote the forgetful functor from $S^{-1}B$-modules to $B$-modules : we have the adjunction $S^{-1}\dashv U$
We have for any $S^{-1}B$-module $M$, $\hom_{S^{-1}B}(S^{-1}\Omega_{B/A}, M) \simeq \hom_B(\Omega_{B/A}, UM)\simeq \mathrm{Der}_A(B,UM)$
Now let $d:B\to UM$ be an $A$-derivation, and define $\tilde{d} : S^{-1}B\to M$ via $\tilde{d}(b/s) = \frac{1}{s^2}(s d(b) - b d(s))$. One easily checks that this is a derivation, and that $\tilde{d}$ agrees with $d$ on $B$.
Conversely given $d : S^{-1}B\to M$, it is clear that by composing with $B\to S^{-1}B$ we obtain a derivation $B\to UM$, and that the two operations are inverse to one another. This is clearly natural in $M$.
Therefore we may continue our string of natural isomorphisms $\mathrm{Der}_A(B,UM)\simeq \mathrm{Der}_A(S^{-1}B,M) \simeq \hom_{S^{-1}B}(\Omega_{S^{-1}B/A},M)$.
By the Yoneda lemma, $\Omega_{S^{-1}B/A}\simeq S^{-1}\Omega_{B/A}$. Now if you track down the proof of the Yoneda lemma and the isomorphisms given here (which are explicit at each step; more or less depending on your definition of $\Omega$), you should recover the canonical maps that your document explicited. 
