Derivation of momentum operator from kinetic energy operator Earlier in class today, my professor explained how to derive the momentum operator. His derivation went like this:
$$\hat{KE} = \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$$
and
$$KE=\frac{P^2}{2m} \Rightarrow P = \sqrt{2mKE}$$
Hence: $$\hat{P} = \sqrt{2m\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}} \Rightarrow \hat{P} = -i\hbar\frac{\partial}{\partial x} \hspace{0.5cm}_{\square}$$
I understood how he got the $-i\hbar$ part, but I can't understand how he wrote $\sqrt{\frac{\partial^2}{\partial x^2}}$ as $\frac{\partial}{\partial x}$.
Is this not a normal derivative operator, hence we can manipulate it as such? Or can $\frac{\partial}{\partial x}$ be operated upon as if it were a number?
I do highly doubt the latter case as it doesn't make any sense to me. For example:
$$y = 3x^2 \Rightarrow \frac{dy}{dx}=9x^2 \Rightarrow \frac{d^2y}{dx^2}=18x$$
yet:
$$\sqrt{18x} \neq 9x^2$$
Am I thinking of this more 'number-based' than it should be? Should $\frac{d}{dx}$ be thought of as a more 'abstract' operator that we can manipulate as we please?
 A: 
$$y = 3x^2 \Rightarrow \frac{dy}{dx}=9x^2 \Rightarrow \frac{d^2y}{dx^2}=18x$$
yet:
$$\sqrt{18x} \neq 9x^2$$

Here you are comparing $\sqrt{\frac{d^2f}{dx^2}}$ to $\sqrt{(\frac{df}{dx})^2}$, which are not the same things. The latter is the first derivative, the former is the square root of the second derivative.


I can't understand how he wrote $\sqrt{\frac{\partial^2}{\partial x^2}}$ as $\frac{\partial}{\partial x}$.

It's a bit sketchy, but I think I can help you feel better about this step.
You should start to think of a symbol like:
$$
\frac{d}{dx}
$$
and an operator that operates on everything to its right (on the written page).
Therefore:
$$
\hat P |f\rangle \to \frac{d}{dx}f(x) \equiv \frac{df}{dx}.
$$
Now call $g(x) = \frac{df}{dx}$. What is $\hat P g$? It is:
$$
\hat P |g\rangle \to \frac{d}{dx}g(x) \equiv \frac{dg}{dx} = \frac{d^2f}{dx^2}\;,
$$
so it is reasonable to symbolically refer to a square root of $\frac{d^2}{dx^2}$ as $\frac{d}{dx}$.


I understood how he got the $=-i\hbar$ part

You should make sure to realize that $+i\hbar$ would, in principle, work just as well and that the negative sign is a convention. It is a conventional sign, but it is so entrenched it is effectively the only sign convention you will ever use. Note, if you change the sign convention here, you must change it a lot of other places, like in the definition of the commutator.
A: Normally (at least as far as I know) $p = \frac {\hbar} {i} \frac{d} {dx}$ is an axiom of quantum mechanics (or more generally you could just postulate that $[x,p] = i \hbar$ and then use the Stone-von-Neumann theorem to show that after a unitary transformation, in position space p becomes $\frac {\hbar} {i} \frac{d} {dx}$) and then you define the Hamiltonian as the Hamilton function known from classical mechanics with every "p" replaced by your p-operator and every "x" replaced with your x-operator (and then in some cases you have to symmetrisize H in order for it to become hermitian).
That being said, I think the key to understanding the derivation by your professor is the definition of the square of an operator. Squaring an operator means applying it two times, so
$A^2 |\psi> = A (A|\psi>)$ and not $A^2 |\psi> = (A|\psi>)^2$. So $(\frac {d} {dx})^2 \psi(x) = \frac {d} {dx} (\frac {d \psi(x)} {dx}) \ne (\frac {d \psi(x)} {dx})^2$. So, also $\sqrt{ \frac{d^2} {dx^2}} = \frac{d} {dx}$.
By the way, you can also define the square root of an operator in general (for example using a power series, though for unbounded operators there are complications with that; another way for self-adjoint operators is to only define how $\sqrt{A}$ acts on their eigenstates and then use that every state can be expressed in those eigenstates), and I am pretty sure that those definitions are consistent with $\sqrt{A^2} = A$ for an operator A.
One last comment: In the derivation given by your professor, you could just as well have defined p as $ + i \hbar \frac {d} {dx}$. One might think that that just shows that it is not sufficient to start with the definition of the free hamiltonian, but that would only be part-right. Because in fact, all of quantum mechanics can be rewritten with i -> -i without losing anything. So, in a way, it is just convention that $p = -i \hbar \frac {d} {dx}$.
