What is $\frac{dx}{d}?$ The operator $\frac{d}{dx}$ is common in calculus to denote a derivative. However, this also begs the question, what is the operator $\frac{dx}{d}$? Is this operator used commonly? If so, what is it called/what does it do? 
I have played aroud with it before, and found a natural way to define it seems to be that $$\frac{dx}{d}\frac{1}{f(x)} := \frac{dx}{df} = \frac{1}{(\frac{df}{dx})}$$
I found also in my own playing around that this could define an odd thing when applying the operator twice:
$$\frac{dx}{d}(\frac{dx}{df}) = \frac{dx}{d\frac{df}{dx}} = \frac{1}{\frac{d\frac{df}{dx}}{dx}} = \frac{1}{\frac{d^2f}{dx^2}}$$
Which would seem to imply a nice notation definition:
$$\frac{dx}{d}(\frac{dx}{df}) := \frac{dx^2}{d^2f}$$
All this is purely my own speculation/invention, of course. I've never heard of any operation like this, and can't find it on the internet, because I don't have a name for it and can't find the notation anywhere. Is this operation already well-defined?
 A: One problem is that viewing $\text{“ } \dfrac {dx} d\text{ ''}$ as an "operator" is that it should potentially be able to operate on something, thus: $$\text{“ } \frac {dx} d f \text{ ''}. \tag 1$$
That makes sense with $\dfrac d{dx}$ because $\dfrac d {dx} f$ means $\dfrac{df}{dx}$.  But nothing analogous can be done with the expression in $(1)$.
A: First of all, your operator needs to act on some space of functions. The obvious restrictions could be non-vanishing functions with non-vanishing derivative. This immediately rules out linear functions and specifically $f\equiv 0$, so your operator won't act on usual vector space of functions, at least not without suitable modifications (perhaps with something involving extension by continuity, as suggested by celtschk in comments).
For example, if we are talking about $C^1$ functions, "non-vanishing functions with non-vanishing derivative" are positive/negative strictly monotonous functions.
However, this still won't be linear (at least not with standard addition). To see this, I will change notation to $f' = \frac{df}{dx}$ to avoid unnecessary clusters.
So, to definition:
$$\frac{dx}d f = \frac{1}{(\frac 1f)'}= \frac{-f^2}{f'}$$
This is homogeneous:
$$\frac {dx}d (af) = \frac {-a^2f^2}{af'} = a\frac{dx}df,$$
but not additive:
$$\frac{dx}df + \frac{dx}dg = \frac{-f^2}{f'} + \frac{-g^2}{g'} = -\frac{f^2g'+f'g^2}{f'g'}=-\frac{(f^2g'+f'g^2)(\frac 1{f'}+\frac 1{g'})}{f'+g'} = -\frac{f^2 + g^2 + \frac{f^2}{f'}g' + f'\frac{g^2}{g'}}{f'+g'} = \frac{dx}d(f+g) - \frac{\frac{f^2}{f'}g' + f'\frac{g^2}{g'}-2fg }{f'+g'}.$$
Notice that $\frac{f^2}{f'}g' + f'\frac{g^2}{g'}\neq 2fg$ in general, for example, take $f(x) = e^x$ and $g(x) = e^{2x}$.
One could hope then that $\frac{dx}{d(f+g)} = \frac{dx}{df} + \frac{dx}{dg}$ but this is even easier to see to fail.
But, there is even more elementary problem with addition. Sum of functions or their derivatives could vanish at some points. To remedy this, we would have to restrict ourselves even more, for example only to positive functions. But for $\frac{dx}d$ to be well defined then, its derivative should be negative. Thus, we could observe strictly decreasing positive functions and let $\frac{dx}d$ to act on those. It will be (positively) homogeneous, but still not additive.
