Why is force described as rate of change of momentum? 
momentum = mass * velocity

Differentiating both sides leads to

force = mass * acceleration

since the mass doesn't participate in the differentiation as it is constant.

Is this a sound definition, or a pointless use of differentiation? Or are there systems (like rockets) where the mass is also a variable wrt time?
 A: The prescription $\mathbf{p} = m\mathbf{v}$ only holds in non-relativistic contexts, while $F = \frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t}$ is true in all contexts.
A: In the simplest case, the system consists of a single object acted on by a constant external force. Since it is only the object's velocity that can change, not its mass, the momentum transferred is
$$Δp = mΔv ,$$
which with the help of a = F/m and the constant-acceleration equation a = Δv/Δt becomes
$$Δp = maΔt$$
$$= FΔt .$$     
Thus the rate of transfer of momentum, i.e. the number of kg·m/s absorbed per second, is simply the external force, 

relationship between the force on an object and the rate of change of its momentum; valid only if the force is constant.

This is just a restatement of Newton's second law, and in fact Newton originally stated it this way.  
A: $\newcommand{\d}[1]{\frac{\mathrm d #1}{\mathrm dt}}$
Well, yes. Rockets work on this very principle. By throwing out some mass in the form of exhaust gases in the opposite direction, the rocket gains momentum (gain since it is thrown out in the opposite direction). This can be equated to a force.
If mass changes, $$F=\d p=m\d v+v\d m=ma+v\d m$$
Note that in empty space, the rocket has no net force on it. So we get $$0=F=m\d v+v\d m=ma+v\d m\implies ma=-v\d m$$
This lets the rocket accelerate without a force. If there was a gravitational force or something, we get:
 $$\frac{Gm_1m_2}{r^2}=m\d v+v\d m=ma+v\d m$$
A: In addition to what everyone else said above, the $F = \frac{dp}{dt}$ relation also makes the link between Newton's third law and conservation of momentum completely obvious, even moreso since there are cases where Newton's third law doesn't really apply, but conservation of momentum is always true.
