TITLE: Why it is absolutely mistaken to cancel out differentials?
QUESTION [7 upvotes]: In many of my physics courses, (don't worry, this is a mathematics question!) My teachers cancel out differentials, and every time, they say: "If a mathematician saw me canceling out this differentials he would get mad, but we are physicists so, there is no problem in doing this." However, still, I haven't seen an acceptable explanation why is this so dramatically wrong. I would appreciate if someone could explain!

REPLY [9 votes]: I agree with the accepted answer: differential notation is a very useful tool for calculations, and in most of the situations where physicists and engineers use it, everything works out fine. That said, I'd like to point out a case where being sloppy with differential notation can lead one to an apparent "proof" of a false statement. I have heard this attributed to Cauchy, although I suspect this attribution is a "mathematical urban legend".
Suppose $f_n$ is a sequence of continuous functions which converges to a function $f$ on $[0,1]$. We ask whether $f$ must be continuous. We write
$$\begin{align}
|f(x+dx)-f(x)| & =|f(x+dx)-f_n(x+dx)+f_n(x+dx)-f_n(x)+f_n(x)-f(x)| \\
& \leq |f(x+dx)-f_n(x+dx)|+|f_n(x+dx)-f_n(x)|+|f_n(x)-f(x)| \end{align}$$
Informally we now think about infinitely large $n$ and infinitely small $dx$. Then all three terms should be infinitely small (the first and third because of convergence and the second because of continuity). So the original side should be infinitely small, and so $f$ should be continuous.
When we formalize the above argument, everything works out provided $f_n$ converge uniformly. But if they converge only pointwise, then this fails: $f_n(x) = x^n$ converges for $x \neq 1$ to $0$ and for $x = 1$ to $1$. Note that unlike uniform convergence, pointwise convergence can be reasonably formulated without developing the axiomatic framework of analysis: all that we need is a way to talk about convergence of sequences of numbers, and a notion of continuity. 
The problem in the argument when we go to approach the problem formally is that we need to pick a single $n$ to control both the first and third terms, and only after doing so do we choose how small $dx$ must be to control the second term. But that means that we choose $n$ before we have chosen $dx$, and without uniform convergence we may need a larger $n$ to control the first term for our chosen $dx$. 
It is hard to even describe this phenomenon in the infinitesimal language! One way to see it is in the hyperreal framework: take $N$ to be an infinite natural number and $dx=1/N$. Then the standard part of $(1-dx)^N$ is not $1$, but rather $e^{-1}$. And now we see the problem: the limit processes $n \to \infty$ and $x \to 1^-$ compete with one another, one trying to pull the result toward $0$ and the other trying to pull the result toward $1$. This effect is missed when we naively say that $n$ is infinite and $dx$ is infinitesimal without saying how they compare to one another.