Is the composition of the differentiating operator commutative? First of all, can I check that $d\over dx$ can be considered an operator, or function (as it says in the title)?
Is the composition of the differentiating operator commutative? In other words, if $u=f(x,y)$ and $x$ and $y$ are dependent on each other, then is this true:
$${d\left({du\over dx}\right)\over dy}={d\left({du\over dy}\right)\over dx}$$
If it is true, then does this mean we can write this instead:
$${d\left({du\over dx}\right)\over dy}={d\left({du\over dy}\right)\over dx}={d^2u\over dxdy}$$
If it isn't true, then which of these is true:


*

*${d^2u\over dxdy}={d\left({du\over dx}\right)\over dy}$

*${d^2u\over dxdy}={d\left({du\over dy}\right)\over dx}$

*${d^2u\over dxdy}$ doesn't mean anything.
And also, if it isn't true, then is there a set of special (and interesting) cases where this holds:
$${d\left({du\over dx}\right)\over dy}={d\left({du\over dy}\right)\over dx}$$
EDIT: To clarify, the case I'm interested in is if $x$ and $y$ are dependent on each other. However, as a second part to this question, I would be interested in what happens if they are independent too.
 A: You have three questions


*

*Is $\frac{d}{dx}$ an operator?


Well, that of course depends on what you mean by $\frac{d}{dx}$. If you mean 
$$ \frac{d}{dx}f=\lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$$
Then if the limit exists for each $x$ you will have "something" that transforms the function $f(x)$ into another function $g(x)$. And that is the definition of "operator."


*If you have $x=g(y)$ and $f(x,g(y))$, then $\frac{d}{dx}\frac{d}{dy}f=\frac{d}{dy}\frac{d}{dx}f$?


The answer to this, I'm afraid, is "in general no." We can check that with an example 
$$f(x,y)=x^2+y^3$$
With $x=g(y)=y^3$ and $y=h(x)=\sqrt[3]{x}$ being your dependence, which is inversible in this case.
$$\frac{d}{dx}f(x,h(x))=\frac{d}{dx}(x^2+x)=2x+1=2y^3+1$$
$$\frac{d}{dy}\frac{d}{dx}f(x,h(x))=6y^2$$
Now let's try to do it in the other order.
$$\frac{d}{dy}f(g(y),y)=\frac{d}{dy}(y^6+y^3)=6y^5+3y^2=6x^{\frac{5}{3}}+3x^{\frac{2}{3}}$$
$$\frac{d}{dx}\frac{d}{dy}f(g(y),y)=10x^{\frac{2}{3}}+2x^{-\frac{1}{3}}=10y^2+2y^{-1}$$
I computed the derivatives in both orders for a general case with the condition that the dependence between x and y is given by the inversible, derivable functions $g$ and $h$, with  $x=g(y)$, $y=h(x)$, and $h=g^{-1}$. The solution is based on the Taylor expansion for the two variables function $f(x,y)$. This expansion requires to know about the partial derivatives of $f(x,y)$. Partial derivative $\frac{\partial f}{\partial x}$ definition is
$$ \frac{\partial f}{\partial x}(x,y)=\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}$$
And identically for $y$, mutatis mutandis. The Taylor expansion of $f(x+\Delta x,y+\Delta y)$ up to the linear term, is then
$$f(x+\Delta x,y+\Delta y)=f(x,y)+\frac{\partial f}{\partial x}(x,y)\Delta x+\frac{\partial f}{\partial y}(x,y)\Delta y+o(|(\Delta x, \Delta y)|)$$
Now, using this expansion and taking the limit 
$$\frac{df}{dx}(x,h(x))=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x,h(x+\Delta x))-f(x,h(x))}{\Delta x}$$
We obtain, 
$$\frac{df}{dx}(x,h(x))=\frac{\partial f}{\partial x}(x,h(x))+\frac{\partial f}{\partial y}(x,h(x))\frac{dh}{dx}(x)$$    
Remembering that $\lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}(x)=\frac{df}{dx}(x)$. The partial derivatives are just functions of $x$ and $y$, and we can calculate further derivatives of them. This step gets longer, but the final result is
$$\frac{d}{dy}\frac{df}{dx}=\frac{\partial^2 f}{\partial x \partial y}(g(y),y)+\frac{\partial^2 f}{\partial x^2}\frac{dg}{dy}(y)+\frac{\partial^2 f}{\partial y^2}\frac{dh}{dx}(g(y))+\frac{\partial^2 f}{\partial x \partial y}(g(y),y)+$$
$$-\left(\frac{dg}{dy}(y)\right)^{-2}\frac{d^2g}{dy^2}(y)\frac{\partial f}{\partial y}(g(y),y)$$
And similarly in the opposite order
$$\frac{d}{dx}\frac{df}{dy}=\frac{\partial^2 f}{\partial y \partial x}(x,h(x))+\frac{\partial^2 f}{\partial y^2}\frac{dh}{dx}(x)+\frac{\partial^2 f}{\partial x^2}\frac{dg}{dy}(h(x))+\frac{\partial^2 f}{\partial y \partial x}(x,h(x))+$$
$$-\left(\frac{dh}{dx}(x)\right)^{-2}\frac{d^2h}{dx^2}(x)\frac{\partial f}{\partial x}(x,h(x))$$
By the way, to complete this derivation you need to know that the derivative of the inverse function is the inverse of the derivative of the function, i.e., $\frac{df}{dg}(x)=\frac{1}{\frac{dg}{dy}(h(x))}$.
Okey, take some time to look at these expresions, you will see that all but the last term are in general identical. Hence the last terms make any difference. For derivatives to be the same you would need
$$\left(\frac{dg}{dy}(y)\right)^{-2}\frac{d^2g}{dy^2}(y)\frac{\partial f}{\partial y}(g(y),y)=\left(\frac{dh}{dx}(x)\right)^{-2}\frac{d^2h}{dx^2}(x)\frac{\partial f}{\partial x}(x,h(x))$$
This doesn't hold in general. You could see, on the other hand, that any linear transformation between $x$ and $y$ will allow commutation, because the second derivatives of $g$ and $h$ would be both zero.


*What is the meaning of $\frac{d^2f}{dxdy}$ in case the operations are noncommutative?


That again is a matter of convention, there is no a good, final answer. However, for me at least, it seems more logical to make 
$$\frac{d^2f}{dxdy}=\frac{d}{dx}\frac{d}{dy}f$$
That also will be consistent with the current convention for partial derivatives in which you have
$$\frac{\partial^2f}{\partial x\partial y}=\frac{\partial}{\partial x}\frac{\partial }{\partial y} f$$
A: The composition of $\dfrac {\partial}{\partial x}$ and $\dfrac {\partial}{\partial y}$ is commutative if holds the Shwartz theorem : https://en.m.wikipedia.org/wiki/Symmetry_of_second_derivatives . Morever the function  $\dfrac {\partial}{\partial x}$ is an operator in in $C^{\infty}(A)$ where $A$ is an open set in $\mathbb {R}^2$ , that is the vector space of the function from $A$ to $\mathbb {R}$ that are differentiabe infinite times.
