TITLE: "Universal" differential identities
QUESTION [6 upvotes]: (This is a cross-post from MSE).
Let $f:\mathbb{R}^d \to \mathbb{R}$ be smooth. The mixed derivatives commute: $f_{xy}=f_{yx}$. This  identity is "universal" in the sense that it holds for any smooth map.
Question:
Are there any universal identities which are not consequences of the commutation of the mixed derivatives?
More explicitly, 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
$$    D_i \circ D_j = D_j \circ D_i \tag{1}.$$
So, 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 $\mathbb{R}^d \to \mathbb{R}^d$, so we can compose operators).
Are there relations in this ring which are not consequences of the fundamental relation $(1)$?.
Edit:
Actually, this is not the "right" algebraic formulation which interests me: I want to be able to talk about relations of the form of $f_x f_y=f_yf_x$ (trivially true) or $f_{xx}=f_yf_{xy}$ (clearly not universal). Thus I need to add multiplication. 
(Without multiplication, there are no additional relations as observed in this answer).
In particular, I am interested to know whether the "Cofactor Lemma" (divergence-free rows, see below) is a "consequence" of the commutation of mixed derivatives (for dimension $d>2$ this involves multiplication, as well as addition and composition).
So, we need to consider some algebraic structure $A$ which is a subset of differential operators (which map $C^{\infty}(\mathbb{R}^d) \to C^{\infty}(\mathbb{R}^d)$), that is "generated" by the $D_i$ via the $3$ operations - addition, composition, and multiplication*. 
(I am not sure if there is a term for such an "algebraic creature", $A$ is a ring w.r.t both operations $(+,\cdot)$ and $(+,\circ)$, but these two "multiplicative" operations have relations, namely $$(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h).$$
Does such a structure have a name?
*By multiplication (as opposed to composition) of operators I mean the following:
$$D_x \times D_y(f)=f_x \cdot f_y \, , \,  D_x∘D_y(f)=f_{xy} \, , \, (D_x \circ D_x) \times D_y(f)=f_{xx}f_y$$ etc. (Here $f$ is a scalar function, to extend the operations to $\mathbb{R}^d$-valued maps, jut act on each component separately.  I am also allowing for the $i$-th component of output to depend on partial derivatives of all components of $f:\mathbb{R}^d \to \mathbb{R}^d$).

The Cofactor Lemma: 
Let $f:\mathbb{R}^d \to \mathbb{R}^d$ be smooth. Then the Cofactor of $df$ has divergence-free rows:
$$\sum_{j=1}^n \frac{\partial(Cof(Du))_{kj}}{\partial x_j} = 0, k=1,...,d.$$
In dimension $d=2$, it 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 $(f_1)_{xy}=(f_1)_{yx},(f_2)_{xy}=(f_2)_{yx}$.
As stated above, for dimension $d>2$ we need multiplication to even phrase the question properly.

REPLY [7 votes]: Let's work in two dimensions for notational simplicity.  We claim that there is no non-trivial polynomial identity of the form
$$ P( f, f_x, f_y, f_{xx}, f_{xy}, \dots ) = 0$$
relating some finite number of derivatives of $f$, for any smooth $f$.  Suppose for contradiction that this is the case.  Taking a functional derivative (i.e. replacing $f$ by $f+ \varepsilon g$ for arbitrary smooth $g$, and then differentiating at $\varepsilon=0$, one obtains a linearised (in $g$) identity of the form
$$ P_f(f,f_x,\dots) g + P_{f_x}(f,f_x,\dots) g_x + \dots = 0.$$
Now, at any point in space, the values of finitely many of the various derivatives $g, g_x, g_y, \dots$ are completely unconstrained (as can be seen by taking $g$ to be a finite Taylor series around that point).  So the only way the above identity can hold for all $f$ and $g$ is if one has
$$ P_f(f,f_x,\dots) = 0$$
$$ P_{f_x}(f,f_x,\dots) = 0$$
etc..  By induction on the degree of $P$, this can only happen if $P_f, P_{f_x}, \dots$ vanish identically, so that $P$ is constant, hence zero, so the identity is trivial.