Ehrenfest's theorem in one guise says (omitting hats for vectors) for suitable operators A
$$\frac{\mathrm d}{\mathrm dt}\langle A\rangle~=~\bigg\langle\frac{\partial A}{\partial t}\bigg\rangle+\bigg\langle\frac{[A,H]}{i\hbar}\bigg\rangle\tag{1}$$
and an "immediate" consequence is that, inserting the position operator x for A in the theorem, using P for the momentum operator,
$$\frac{\mathrm d}{\mathrm dt}\langle x\rangle~=~\bigg\langle\frac{P}{m}\bigg\rangle.\tag{1a}$$
Ignoring the left side of (1) I reasoned (omitting test function in expectation values)
\begin{align} \bigg\langle\frac{\partial x}{\partial t}\bigg\rangle + \bigg\langle\frac{[x,H]}{i\hbar}\bigg\rangle &=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle + \bigg\langle\frac{xH-Hx}{i\hbar}\bigg\rangle \\ &=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle + \bigg\langle\frac{xH}{i\hbar}\bigg\rangle~-~\bigg\langle\frac{Hx}{i\hbar}\bigg\rangle \\ &=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle + \bigg\langle\frac{xH}{i\hbar}\bigg\rangle~-~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle \end{align} and so
$$\frac{\mathrm d}{\mathrm dt}\langle x\rangle~=~ \bigg\langle\frac{xH}{i\hbar}\bigg\rangle,\tag{2}$$
which I can convince myself is true.
On the other hand, using the first relation in this problem set (PDF):
$$[x,H]=\frac{i\hbar}{m}P, $$ we get immediately that
$$\frac{\mathrm d}{\mathrm dt}\langle x\rangle~=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle+\bigg\langle\frac{[x,H]}{i\hbar}\bigg\rangle$$
$$\frac{\mathrm d}{\mathrm dt}\langle x\rangle~=~\bigg\langle\frac{\partial x}{\partial t}\bigg\rangle+\bigg\langle\frac{P}{m}\bigg\rangle\tag{3}$$
but then from (1a) that $\bigg\langle\partial x/\partial t\bigg\rangle~=0,$ which bothers me a little but doesn't seem to contradict anything above directly.
My question is whether the conclusions (2), (3) in these two calculations are right and if not where I went astray.
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