Product of two differentials (infinitesimals) In a physics derivation today my lecturer ended up with two differentials multiplied together, like this term in a longer summation:
$p^2 q^2 \mathrm{d} p ~ \mathrm{d} q$
And then he said this term becomes zero because the two differentials are multiplied together, and can be removed. Is there a calculus justification for this? It doesn't make sense to me, but we got the right result...
 A: Often when physicists write $\mathrm{d}p$ on its own, they view it as some very small quantity representing some change in $p.$ For example, if $\dfrac{\mathrm{d}p}{\mathrm{d}t} =5$ then they might write $\mathrm{d}p = 5\mathrm{d}t $ and interpret that as "a small change in time causes a small change in $p$ five times greater". Also, when they ignore terms with multiple differentials, they are using their intutition that when we are already dealing with very small quantities, the terms that are products of such small quantities are an entire order of magnitude smaller and thus, negligible.
They do this to avoid repeating the same steps of formalism to reach the same result more rigorously, which would be to use $\Delta t = t-t_0$ to represent some change in time, $\Delta p = p(t) - p(t_0)$ to represent the corresponding change in $p,$ do all their working like this, and then finally taking the limits like $\dfrac{\mathrm{d}p}{\mathrm{d}t} = \displaystyle\lim_{\Delta t \to 0} \frac{ \Delta p}{\Delta t} $ so all the $\Delta$'s become $\mathrm{d}$'s. 
So now suppose we were doing it properly and we have gotten to this:
$$\Delta y = 52 \Delta p + \pi \Delta q + p^2 q^2 \Delta p \Delta q.$$
Divide through by $\Delta t$:
$$\frac{\Delta y}{\Delta t} = 52 \frac{\Delta p}{\Delta t} + \pi \frac{\Delta q}{\Delta t} + p^2 q^2 \frac{\Delta p}{\Delta t} \Delta q.$$
Now when we take the limit as $\Delta t \to 0$ we get $$\dfrac{\mathrm{d}y}{\mathrm{d}t} = 52\dfrac{\mathrm{d}p}{\mathrm{d}t} +\pi \dfrac{\mathrm{d}q}{\mathrm{d}t}. $$ 
The phenomenon you observed occurs - the final result makes it appear that only the terms with a single differential mattered while terms with two or more differentials vanish. That is because in this limit, all $\Delta$ terms go to $0$ but the ones written above as ratios tend towards derivatives, so in the third term: 
$$p^2 q^2 \frac{\Delta p}{\Delta t} \Delta q\to p^2q^2 p'(t) \cdot 0 =0.$$
So the reason you can ignore them is because when you do the background formalism with limits, the terms with two or more differentials will go to $0$ in the end anyway.
