= Solution
Differentiating the <expectation value> and using $\partial_t|\psi\rangle=-i\hat H|\psi\rangle/\hbar$ and $\partial_t\langle\psi|=i\langle\psi|\hat H/\hbar$ gives
$$
\boxed{\frac{d\langle\hat A\rangle}{dt}=\frac i\hbar\langle[\hat H,\hat A]\rangle+\left\langle\frac{\partial\hat A}{\partial t}\right\rangle.}
$$
The second term means explicit operator time dependence, distinct from the state's evolution.
The <commutator product rule> follows by adding and subtracting $\hat B\hat A\hat C$:
$$
[\hat A,\hat B\hat C]=\hat A\hat B\hat C-\hat B\hat C\hat A=[\hat A,\hat B]\hat C+\hat B[\hat A,\hat C].
$$
For the \b[<mathematical induction> step], let $\hat D=[\hat A,\hat B]$ and use $[\hat A,\hat D]=0$. The case $n=1$ is immediate. If $[\hat A^m,\hat B]=m\hat D\hat A^{m-1}$, then
$$
[\hat A^{m+1},\hat B]=\hat A^m\hat D+[\hat A^m,\hat B]\hat A=(m+1)\hat D\hat A^m.
$$
Thus the claimed identity holds for every positive integer. Commutation of $\hat B$ with $\hat D$ is not needed for this particular step. This is the <central-commutator differentiation rule>. Termwise application to a convergent <power series>, on a common domain where the operations are justified, gives
$$
\boxed{[F(\hat A),\hat B]=\sum_{n\geq1}na_n\hat D\hat A^{n-1}=\hat D F'(\hat A).}
$$
For <position> and <momentum>, $[\hat x,\hat p]=i\hbar\hat I$. Hence $[\hat p^2,\hat x]=-2i\hbar\hat p$, $[V(\hat x),\hat x]=0$ and $[V(\hat x),\hat p]=i\hbar V'(\hat x)$. Substitution gives \b[<Ehrenfest theorem>]:
$$
\boxed{\frac{d\langle\hat x\rangle}{dt}=\frac{\langle\hat p\rangle}{m},\qquad\frac{d\langle\hat p\rangle}{dt}=-\langle V'(\hat x)\rangle.}
$$
The mean <momentum> changes by the mean <force>. It does not follow for a general nonlinear <Potential energy> that $\langle V'(\hat x)\rangle=V'(\langle\hat x\rangle)$: a closed classical trajectory for the mean <position> requires an additional condition, such as a quadratic <Potential energy> or a sufficiently localized <wave packet>.
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