I was trying to prove it 2 ways,
1st way: $$p=mΔV$$ $$a=ΔV/t$$ $$ΔV=at$$ $$p=m*a*t$$
Remember $$F=ma$$ $$F(t)=p$$
The derivative of momentum just gives us the "regular force" since b4 that momentum = force as a function of time. NOT SURE IF THIS PART IS CORRECT thus, $$dp/dt = F$$
2nd way: if mass was constant $$p = mv$$ $$dp/dt = m(dv/dt) + v(dm/dt)$$ $$dp/dt = ma + 0, dp/dt = f$$
$F = \mathrm{d}p/\mathrm{d}t$ follows directly from $F=ma$ and the definition of mechanical momentum $p=mv$. $F=ma$ is validated ultimately by experiment. That's all that's needed to say. Newton' second law in any form is valid only for constant mass systems. (For some reason that's a theme here lately.)
The first proof isn't quite right for what you're trying to do.
For a constant mass the impulse momentum theorem states that the change in the momentum is equal to the impulse delivered to the object by the forces action on it. If we consider changes which occur over a very short period of time we can write the change in the momentum as,
$$ \Delta \vec{p} = m \Delta v,$$
and the impulse as.
$$ \vec{J} = \vec{F} \Delta t $$
Newtons second law states that $\vec{F}=m \Delta \vec{v}/ \Delta t$, substituting this into our expression for $\vec{J}$ we get,
$$ \vec{J} = \left( \Delta \vec{v}/ \Delta t \right) \Delta t = m \Delta \vec{v} = \Delta \vec{p}$$
Now to extend the result for a force applied over a finite time interval of length $T$ we integrate to get the above,
$$ \Delta p = J = \int_0^T F(t) dt $$
Force is defined as $dP/dt=F$.
It is equally valid for both constant and varying mass systems. just that $F=ma$ is insufficient in varying mass system. You need to add the thrust force ( $vdm/dt$) also.
The general formula is $dp/dt=d(mv)/dt=vdm/dt+ma$ (using product rule).
But yes at the core level it is simply defined as such.
