Hmmm I realized I have made a mistake in what I am saying. What I am saying is subtle so I'm having trouble articulating it, in that I do believe:
\[\frac{d \langle x \rangle}{dt} = \langle \frac{dx}{dt} \rangle \]
However I don't believe that these two expressions that result within the integral are necessarily identical. Although now I realize my argument is sort of going back to where I originally found it which is not entirely within the realm of mathematics.
\[\frac{d}{dt} \int_{-\infty}^\infty x \Psi^*(x,t) \Psi(x,t) dx = \int_{-\infty}^\infty x \Psi^*_t(x,t) \Psi_t(x,t) dx\]
I don't exactly believe the fact that we are essentially dropping out the term \[\frac{dx}{dt}=0\] in differentiation under the integral sign since it seems to me that position could be thought of as depending on time... Although I understand there's the ensemble interpretation, I just don't feel convinced that:
"The derivative of the expectation value of position is the expectation value of velocity"
These seem to me to be two different things, the first being the velocity of the average particle and the second is the average velocity. So while these integrals for expectation values align in the same way that:
\[\langle v \rangle = \frac{1+2+3}{3}=2\]
\[\frac{d}{dt} \langle x \rangle = \frac{2+2+2}{3}=2\]
I guess what I'm saying is I don't think the momentum operator produces the momentum, even though it ends up in the integral evaluating to the same expected value.