= Solution
For $H=p^2/(2m)+V(x)$ and a time-independent operator $O$, differentiate its <expectation value> and use the <Schrödinger equation> and its adjoint. Moving $H$ across the inner product by integration by parts gives
$$
\frac{d}{dt}\langle O\rangle=\frac{i}{\hbar}\langle HO-OH\rangle=\frac{i}{\hbar}\langle[H,O]\rangle.
$$
Work with wavefunctions in the displayed operators' domains; the vanishing at infinity removes the associated boundary terms. This is the <Ehrenfest theorem> derived from time evolution.
The <canonical commutation relation> follows directly from differentiating $x\psi$: $[p,x]\psi=-i\hbar\psi$. Thus $[p^2,x]=p[p,x]+[p,x]p=-2i\hbar p$, while $[V(x),x]=0$. It follows that
$$
\boxed{\frac{d}{dt}\langle x\rangle=\frac{\langle p\rangle}{m}.}
$$
For the momentum equation, apply both products to a test wavefunction:
$$
[V,p]\psi=-i\hbar V\psi'+i\hbar(V\psi)'=i\hbar V'\psi.
$$
The kinetic term commutes with $p$, so $[H,p]=i\hbar V'$. Therefore
$$
\boxed{\frac{d}{dt}\langle p\rangle=-\langle V'(x)\rangle.}
$$
These are the <classical equations from Ehrenfest theorem> for position and momentum expectations; the force expectation need not equal the force evaluated at the expected position.
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