So this question is a direct follow on from my last question
I am trying to get an expression for $\frac{d\left<p\right>}{dt}$.
I have my working out below, which was thanks to the Anonymous Physicist's help, (not to mention all the patient people in the mathematics chatroom) I have got to this expression: $$ i\hbar \left( \int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx + \int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx \right). $$
And now I'm stuck. What can I do about $$ \int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx $$
In particular how to deal with $$ \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x}? $$ With the $\partial t$ in play I don't know how solve the integral...
As before, I really would appreciate hints to get me moving in the right direction more than an outright solution.
My working so far
We are given $\left<p\right>$ in the text, which is determined by computing $\frac{d\left<x\right>}{dt}$ and multiplying by $m$.
Lets just start with what we know: $$ \left<p\right> = \int{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx. $$
So taking the time derivative, $$ \frac{d\left<p\right>}{dt} = \frac{d}{dt}\int{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx. $$
We probably should at this point pause, and listen to Ted, (and RobJohn). So take out the constants before going any further. We get, $$ -i\hbar \frac{d}{dt}\int\limits_{-\infty}^{+\infty}{ \left( \Psi^* \left[ \frac{\partial}{\partial x} \right] \Psi \right) }dx. $$
Bringing the time derivative into the integral, gives us $$ -i\hbar \int\limits_{-\infty}^{+\infty}{\frac{\partial}{\partial t}\Psi^*\left[\frac{\partial}{\partial x}\right]\Psi}dx. $$
Then differentiating with the product rule yields, $$ -i\hbar \int\limits_{-\infty}^{+\infty}{ \frac{\partial \Psi^*}{\partial t} \frac{\partial \Psi}{\partial x}+ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx. $$
Now we have two terms in our integral, let's split em up. Let's consider the left handside first, we have $$ \int\limits_{-\infty}^{+\infty}{ \frac{\partial \Psi^*}{\partial t} \frac{\partial \Psi}{\partial x} }dx. $$
By Schrödinger this can become, $$ \int\limits_{-\infty}^{+\infty}{ \left(\frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar}V \Psi^* \right) \frac{\partial \Psi}{\partial x} }dx. $$
Expand to get, $$ \int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x}+ \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx. $$
So, just to review, our full problem can be written as, $$ i\hbar \left( \int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx + \int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx \right). $$
