In the expression of time derivative of expectation value of position, it is mentioned in book (Introduction to Quantum Mechanics by DJ Griffith) that inside the integral, the differentiation with respect to position $x$ will be zero, so the time derivative only applies on probability density.
But I don't get the point, why inside the integral, time derivative of position should be zero?
Now as time goes on, $\langle x \rangle$ will change (because of the time dependence of $\Psi$), and we might be intered in knowing how fast it moves. Referring to Eequations 1.25 and 1.28, we see that $$ \frac{d\langle x \rangle}{dt} = \int x \frac{\partial}{\partial t} \left \lvert \Psi \right \rvert^2 dx = \frac{i \hbar}{2m}\int x \frac{\partial}{\partial x}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \, dx \, . $$