How to derive the fact that $p\sim d/dx$ and $H\sim d/dt$ from classical mechanics? I am trying to understand Noether's conserved quantities to shifts in time and or position.  I have seen the derivation of the operators for Schrodinger's equation but not for classical mechanics.
Is it true or perhaps obvious that the derivative of the action $S=∫(KE-PE)dt$ with respect to time is Energy since it is the $d/dt$ of a $∫dt$ that contains energy?  Also is it equally obvious that the derivative of the action with respect to position, is the momentum $dL/dx=$ momentum?  Or does it require some extra steps and logic to derive the fact that $d/dx=$ momentum and $d/dt=$ Hamiltonian?
 A: These observables-are-derivatives equations carelessly obscure what's really happening, viz.$$\langle x|\hat{p}_k|\psi\rangle=-i\hbar\frac{\partial}{\partial x_k}\langle x|\psi\rangle,\,\hat{H}|\psi\rangle=i\hbar\frac{\partial}{\partial t}|\psi\rangle$$or something like that. Since these statements are in quantum-mechanical Hilbert space language, they cannot be derived in classical mechanics. Instead, quantum mechanics is constructed to represent certain truths about classical mechanics in these equations. The Poisson brackets $\{x_j,\,p_k\}=\delta_{jk}$ give rise to $[x_j,\,p_k]=i\hbar\delta_{jk}\Bbb I$ ($\Bbb I$ the Hilbert space's identity operator) because momentum and the Hamiltonian are the generators of infinitesimal space and time translations (see here and here, respectively).
A: The correspondence with classical mechanics is best seen in the Heisenberg picture by acting on operators $\hat{f}(\hat{x},\hat{p},t)$ rather than on kets $|\psi\rangle$.
First of all, recall that the commutator $\frac{1}{i\hbar} [\cdot,\cdot]$ corresponds to the Poisson bracket $\{\cdot,\cdot\}$, cf. e.g. this Phys.SE post.
Now let us answer OP's title question:

*

*Differentiation of the operator $\hat{f}$ wrt. time $t$ is given by Heisenberg's EOM
$$ \frac{d\hat{f}}{dt}~=~ \frac{1}{i\hbar} [\hat{f},\hat{H}]+\frac{\partial \hat{f}}{\partial t}, \tag{1Q}$$
which corresponds to Hamilton's EOM
$$ \frac{df}{dt}~=~\{f,H\}+\frac{\partial f}{\partial t}. \tag{1C}$$


*Differentiation of $\hat{f}$ wrt. $\hat{x}$ is given by
$$  \frac{1}{i\hbar} [\hat{f},\hat{p}], \tag{2Q}$$
which corresponds to
$$ \frac{\partial f}{\partial x}~=~\{f,p\}.\tag{2C} $$
