.(This is a cross-post from MSE)

Consider the following well-known fact:

Let $f:\mathbb{R}^2 \to \mathbb{R}$ be smooth. Then the mixed partial derivatives commute: $f_{xy}=f_{yx}$. This identity is "universal" in the sense that it holds for any smooth map, no matter what are its properties or any special features it may or may not posses (similarly for maps $\mathbb{R}^d \to \mathbb{R}$).

Question:

Are there any universal identities (satisfied by all smooth maps) which are not consequences of the commutation of the mixed derivatives?

One way to phrase this is mentioned here:

Let $D_i$ be the differential operator which takes the partial derivative with respect to $x_i$. The symmetry can be written as an algebraic statement:

$$ D_i \circ D_j = D_j \circ D_i \tag{1}$$

From this relation it follows that the ring of differential operators with constant coefficients, generated by the $D_i$, is commutative.

(When we choose the domain for these operators to be the space of smooth maps $\mathbb{R}^d \to \mathbb{R}^k$).

Are there relations in this ring which are not consequences of the fundamental relation $(1)$?.

It's not always trivial to see explicitly if a relation comes from the symmetry:

For example, there is the following universal result:

Let $f:\mathbb{R}^d \to \mathbb{R}^d$ be smooth. Then the Cofactor of $df$ has divergence-free rows:

$$\sum_{j=1}^n \frac{\partial(Cof(Du))_{kj}}{\partial x_j} = 0$$

For the case $d=2$, It indeed reduces to relation $(1)$:

Given $A= \begin{pmatrix} a & b \\\ c & d \end{pmatrix}$, $\operatorname{Cof}A= \begin{pmatrix} d & -c \\\ -b & a \end{pmatrix}$, so

$$ df= \begin{pmatrix} (f_1)_x & (f_1)_y \\\ (f_2)_x & (f_2)_y \end{pmatrix}, \operatorname{Cof}df= \begin{pmatrix} (f_2)_y & -(f_2)_x \\\ -(f_1)_y & (f_1)_x \end{pmatrix} .$$

We see that $\operatorname{div} (\operatorname{Cof}df)=0$ is equivalent to $(f_1)_{xy}=(f_1)_{yx},(f_2)_{xy}=(f_2)_{yx}$.

I guess this also hold for $d>2$ but this seems less transparent.