**Proof.**
To show the equality of (1) it is enough to show that any $A$-derivation $D : B \to M$ annihilates the elements $\varphi (s)^{-1}$. This is clear from the Leibniz rule applied to $1 = \varphi (s) \varphi (s)^{-1}$. To show (2) note that there is an obvious map $S^{-1}\Omega _{B/A} \to \Omega _{S^{-1}B/A}$. To show it is an isomorphism it is enough to show that there is a $A$-derivation $\text{d}'$ of $S^{-1}B$ into $S^{-1}\Omega _{B/A}$. To define it we simply set $\text{d}'(b/s) = (1/s)\text{d}b - (1/s^2)b\text{d}s$. Details omitted.
$\square$

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