I start by showing how I tried to obtain the position operator analogously to how
the momentum operator is obtained:

If we differentiate the wave function in one dimension $\Psi(x,t) = e^{i(px-Et)/\hbar}$, with respect to x:

$$
\frac{\partial}{\partial x}\Psi(x,t) = \frac{ip}{\hbar}~\Psi(x,t)
\tag1
$$

from which we get the momentum operator: $\hat{p}=-i\hbar\frac{\partial}{\partial x}$

But suppose I differentiate $\Psi$ with respect to momentum:

$$
\frac{\partial}{\partial p}\Psi(x,t) = \frac{i}{\hbar}(x-\frac{p}{m}t)~\Psi(x,t)
\tag2
$$

which gives $x-\frac{p}{m}t = -i\hbar\frac{\partial}{\partial p}$

Now, setting $t=0$, I suppose we would get the operator $$\hat x\_0 = -i\hbar\frac{\partial}{\partial p}$$ Which would be something like an operator for initial position. But it doesn't look right, because I know that the position operator does not have a minus sign. And is there even such a thing as "initial position operator"?

So, what is wrong with this? The reason I'm asking is because I want to show that the expectation value for position satisfies the following relation (in accordance with the correspondance principle):

$$\left<x\right> = \left<x\right>\_{t=t\_0}+\frac{\left<p\right>}{m}(t-t\_0)$$

and it was given as a hint to start as in $(2)$ and take it from there. I sort of know what to do, but the minus sign in the position operator confuses me.

The hint also suggests $(2)$ should lead to $x = i\hbar\frac{\partial}{\partial p} + \frac{p}{m}t$, but somehow the minus sign isn't there.