When are differential operators not equivalent to variables? Previously I asked how to integrate:
$$\frac{dy}{dx}=\frac{x}{2y}$$   
Posters said that I could simply move the $2y$ and $dx$ over, essentially treating the $dx$ as a regular variable. 
I wonder then why we cannot always treat differential operators as variables? I was always taught not to think of them as variables. Why not? 
 A: Strictly speaking, no, you cannot "move the $dx$ over". $\frac{dy}{dx}$ is not a fraction, it is a function, a full symbol in its own right, and the numerator and denominator are as inseparable as the l and the n in $\ln(x)$. However, you may move the $2y$ over, and then integrate both sides with respect to $x$ (if the function on the left-hand side is equal to the function on the right-hand side, then their integrals must also agree, up to a constant term). This gives
$$
\int 2y\frac{dy}{dx} \,dx = \int x\,dx
$$
Then you might recognize that what you have on the left-hand side is the chain rule for integration, which says that
$$
\int 2y \frac{dy}{dx}\,dx = \int 2y\,dy
$$
This allows you to solve the differential equation simply by calculating the integrals. 
That being said, note that you would've gotten the same answer had you treated $dx$ as a variable and moved it to the other side. This is one of the reasons why writing derivatives as $\frac{dy}{dx}$ is so popular: even though it isn't a fraction, and the "numerator" and "denominator" aren't really terms by themselves, you still get the correct answer if you pretend that they are.
A: The method of separation of variables is nothing more than an application of the chain rule, so you can avoid the symbols $dx$ and $dy$ altogether:
If $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{f(x)}{g(y)}$, then
$\frac{\mathrm{d} y}{\mathrm{d} x}g(y)=f(x)$.
If $g$ has an antiderivative, $G$ then $\frac{\mathrm{d} G}{\mathrm{d} y}=g(y)$ and we get 
$\frac{\mathrm{d} y}{\mathrm{d} x}\frac{\mathrm{d} G}{\mathrm{d} y}=f(x)$.
The LHS is $(G\circ y)'(x)$ from which it follows that
$(G\circ y)(x)=\int f(x)dx.\ $ (Note the antiderivative is taken wrt $x$ on $both$ sides). 
On the other hand, $(G\circ y)(x)=G(y(x))=G(y)=\int g(y)dy$. 
Putting this all together, we have the desired result 
$\int g(y)dy=\int f(x)dx.$
Remark: I guess to treat $dx$ and $dy$ rigorously, it is best to regard them as linear functions on tangent spaces.
