Let $B$ be an $A$-algebra and let $d:B\rightarrow \Omega_{B/A}$ be its $B$ module of Kähler differentials relative to $B$. By an explicit construction of $\Omega_{B/A}$, I know that the $A$-module generated by the image of $d$ is the entire $\Omega_{B/A}$.
How could I prove this by using only the universal property of $d:B\rightarrow \Omega_{B/A}$?
Let $M\subset\Omega_{B/A}$ be the submodule generated by the image of $d$, let $i:M\to \Omega_{B/A}$ be the inclusion map, and let $d':B\to M$ be $d$ with its codomain restricted to $M$ (so $d=id'$). By the universal property of $d$, there is a unique homomorphism $f:\Omega_{B/A}\to M$ such that $fd=d'$. Composing with $i$ we have $$ifd=id'=d.$$ But now $if$ and $1_{\Omega_{B/A}}$ are two different homomorphisms $g:\Omega_{B/A}\to\Omega_{B/A}$ such that $gd=d$, and so by the uniqueness part of the universal property of $d$, we have $if=1_{\Omega_{B/A}}$. This implies $i$ is surjective so $M$ is all of $\Omega_{B/A}$.
(Alternatively, you can observe that $d'$ has the same universal property as $d$ (think about why!) which means $i$ is an isomorphism, by the usual proof that universal objects are unique up to isomorphism.)