Confusion about derivative notation I have always been confused about Leibniz notation. Not the notation itself, but the fact that it treats the differential operators ($d$, $\partial$) as being multipliable. The most famous example would probably be the Schrödinger equation, which if often denoted something like this:
$$E\psi=\left(-\frac{h^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right)\psi=-\frac{h^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V(x)\psi$$
The problem with this is that is makes use of multiplication to expand the expression, $(a+b)c=a*c+b*c$, which would imply that $\partial *f=\partial f$.
Personally, I would define the differential operators as functions:
$$d(f)=\lim_{h\rightarrow \infty}{\frac{f(x+h)-f(x)}{h}}$$
So the above statement would make no sense at all. First I just accepted it as mathematical laziness, but then I stumbled upon this monstrosity:

Observe that
  $$\left(v^2\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial t^2}\right)y=0$$
  can be factored as (which is what you probably mean by "squaring" in the question)
  $$\left(v\frac{\partial}{\partial x}+\frac{\partial}{\partial t}\right)\left(v\frac{\partial}{\partial x}-\frac{\partial}{\partial t}\right)y=0$$

What is happening here???

$-\frac{h^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)$
 is the sum of two operators.  $V(x)$ is understood to be the multiplication operator, that is, the operator $V(x)$ applied to $\psi$ is the product of the functions $V(x)\psi(x)$.  The sum of operators is defined as $(A+B)\psi = A\psi+B\psi$.
