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Here is a sketch of a proof that $d (e^x) \neq e^x dx$ in the Kahler differentials of $C^{\infty}(\mathbb{R})$. More generally, we should be able to show that, if $f$ and $g$ are $C^{\infty}$ functions with no polynomial relation between them, then $df$ and $dg$ are algebraically independent, but I haven't thought through every detail.

Choose any sequence of points $x_1$, $x_2$, in $\mathbb{R}$, tending to $\infty$. Inside the ring $\prod_{i=0}^{\infty} \mathbb{R}$, let $X$ and $e^X$ be the sequences $(x_i)$ and $(e^{x_i})$. Choose a nonprincipal ultrafilter on the $x_i$ and let $K$ be the corresponding quotient of $\prod_{i=0}^{\infty} \mathbb{R}$.

$K$ is a field. Within $K$, the elements $X$ and $e^X$ do not obey any polynomial relation with real coefficients. (Because, for any nonzero polynomial $f$, $f(x,e^x)$ only has finitely many zeroes.) Choose a transcendence basis, $\{ z_a \}$, for $K$ over $\mathbb{R}$ and let $L$ be the field $\mathbb{R}(z_a)$.

Any function $\{ z_a \} \to L$ extends to a unique derivation $L \to L$, trivial on $\mathbb{R}$. In particular, we can find $D:L \to L$ so that $D(X)=0$ and $D(e^X) =1$. Since $K/L$ is algebraic and characteristic zero, $D$ extends to a unique derivation $K \to K$. Taking the composition $C^{\infty} \to K \to K$, we have a derivation $C^{\infty}(\mathbb{R}) \to K$ with $D(X)=0$ and $D(e^X)=1$. By the universal property of the Kahler differentials, this derivation factors through the Kahler differentials. So there is a quotient of the Kahler differentials where $dx$ becomes $0$, and $d(e^x)$ does not, so $dx$ does not divide $d(e^x)$.

I'm traveling and can't provide references for most of the facts I am using aobut derivations of fields, but I think this is all in the appendix to Eisenbud's Commutative Algebra.

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There is a discussion of this issue at the $n$-category cafe. I'd encourage people who were interested in this question to head over there and see if they can lend some insight.