Second Order Term Ignored [Tsiolkovsky's Rocket Equation] I am working on the derivation of Tsiolkovsky's Rocket Equation, so far I have started with newton's law of conservation of momentum
mv = (m+dm)(v+dv) 
Where
Rocket has mass “m”
Particle ejected from the engine has mass “dm”
Exhaust velocity is defined by “Ve”
Change of the velocity of the rocket is defined by dv
Rocket has a current velocity of v
After distributing and simplification I got
1 0 = mdv + dmdv + dmVe
I was rather stuck at this point so I looked at some other articles on the topic and found that the next step would be to "ignore dmdv as a second order term", so the equation would be:
2 0 = mdv + dmVe
I know quite a bit of Calculus, but I must have forgotten (or was never taught) Second Order Terms. Can someone explain how one can progress from step 1 to step 2?
Thanks!
 A: The idea is that $dm $ and $dv $ are both small. When we multiply small things together, they just get smaller, so $dm\,dv $ is really small. Since this term is so small, we can make the problem simpler by ignoring it without sacrificing accuracy too much. In general, an $n^\text {th} $ order term is any term with $n $ small things multiplied together.
A: Been working through this myself, and I found another solution that avoids the product of "infinitesimals" altogether.
I was having similar issues with this equation because the derivation gets complicated depending on how you define the initial mass (assume there are no external forces in either case, and we always replace the exhaust velocity $V_e$ in the observer frame with the exhaust velocity in the rocket frame, thus $V_e = v - v_e$).

Version 1:

Initial mass: $dm + m$
$p_1 = (m + dm)v$
$p_2 = dm(v - v_e) + m(v + dv)$
$p_1 = p_2$
$mv + vdm$ = $vdm - v_edm + mv + mdv$
$v_e\frac{dm}{m}$ = $dv$
Which is easily integrated, and conveniently doesn't contain the $dmdv$ term.

Version 2:

Initial mass: $m$
$p_1 = mv$
$p_2 = dm(v - v_e) + (m - dm)(v + dv)$
$p_1 = p_2$
$mv = vdm - v_edm + mv + mdv - vdm - dmdv$
Which is very similar but has the extra $-dmdv$.
It seems weird to me that we sometimes have to ignore this small error and sometimes it's not even there in the first place.
I got a bit of insight when I tried to separate the variables, although my calc isn't strong enough to make it rigorous:
$v_e\frac{dm}{m} = dv(1 - \frac{dm}{m}) = dv(\frac{m - dm}{m})$
$v_e\frac{dm}{m - dm} = dv$
This feels like an affirmation of the fact that it's so small in comparison: the sum m - dm is basically m, since the latter is so much smaller than the former.
To complete the insight, consider this much simpler equation that you've done a hundred times:
$y = x^2$
$$\frac{dy}{dx} = \lim_{h\to0}\frac{(x+h)^2 - x^2}{h} = \lim_{h\to0}\frac{h^2 +2xh}{h} = \lim_{h\to0}h +2x = 2x$$
The trick here was moving things around until the division by zero went away.
When we started our derivation, we used $dm$ and $dv$, but really we were using that as shorthand for $\Delta m$ and $\Delta v$ and then taking the limit.
So going back to
$v_e\frac{dm}{m - dm} = dv$
And replacing it with
$v_e\frac{\Delta m}{m - \Delta m} = \Delta v$
We can see that there would be division no by zero using $\Delta m = 0$ in the denominator. Again, I can't really connect the dots back to fundamentals, but this makes me think that this term can actually be safely considered zero with sufficient calculus rigor, and that it exists in the first place because of my use of differential terms without fully understanding their relation back to derivatives/limits.
