The following question is asked without malicious intentions - it's not intended as a flamebait!
In my physics textbooks (Young & Freedman in particular) I have often seen derivations of equations that use multiplication of, say, $dx$ on both sides, and then integrating.
Taking courses both at the math and physics department I've noticed that real mathematicians often frown at this. Why is that?
Suppose you have the equation $$f(x) = g(x).$$ You could "multiply both sides by $dx$" $$f(x) \, dx = g(x) \, dx$$ and then integrate over some interval $$\int_a^b f(x) \, dx = \int_a^b g(x) \, dx.$$ However, unless you have learned about differential forms, the second equation above is meaningless. This can easily be avoided by simply integrating both sides of $f(x) = g(x)$ with respect to $x$ just as you learned in calculus.
Now consider the differential equation $$f(y) \frac{dy}{dx} = g(x).$$ Here, many people will "multiply by $dx$" to get $$f(y) \, dy = g(x) \, dx$$ and then integrate both sides. Again, without the machinery of differential forms, this statement is meaningless. Here we can avoid this by integrating the original equation with respect to $x$: $$\int f(y) \frac{dy}{dx} \, dx = \int g(x) \, dx$$ or to use different notation, $$\int f(y(x))y'(x)\, dx = \int g(x) \, dx.$$ Then as long as certain hypotheses are satisfied, the change-of-variables (substitution) theorem says that this is just $$\int f(y) \, dy = \int g(x) \, dx$$ so we are back to where the first method lead us.
The point of this is that manipulations involving "multiplying by $dx$" are usually shorthand for some more "rigorous" method. However, if you are aware of (and comfortable with) the mathematics going on behind the scenes, then there is no loss of rigor. "Real mathematicians" don't frown upon using these kinds of manipulations themselves, they just frown upon students using them without knowing why they work.
Because it is considered as non rigorous. Calculating with infinitesimals is something physicists and mathematicians have done since at least the time of Fermat, even long before that. But it always felt awkward to work with infinitesimals as they seemed to lead to contradictory statements if not manipulated carefully.
In the 19th century, under the impulse of Cauchy, calculus started to be formalized and the calculus of infinitesimals was replaced by the more precise and rigorous $\epsilon-\delta$ methodology. Physicists didn't take much notice however since calculating with differentials still worked and was more practical.
In the 1960's however, Abraham Robinson developed a rigorous approach to manipulating infinitesimals now called non-standard analysis. At the time it was controversial, nowadays, I think it is more accepted. That doesn't mean physicists really care that much, but in a way, Robinson vindicated their approach. Although you still need some work to make things rigorous.
I think the earlier answers to this question missed the mark.
In physics, you never know exactly what you're studying. There's always uncertainty in your knowledge of the initial conditions of a system, and even if you could plug exact numbers into your model, the model itself is just an approximation.
Increasing the precision of numerical integration will yield a more accurate approximation up to a point, but beyond that point (somewhere around the atomic level, say), more precision makes the model worse. Fortunately, the result often doesn't drift too far from the right answer even in the infinitesimal limit, which means calculus is useful. But it's not correct. We know how every continuum model breaks down except for the spacetime continuum, and there are plenty of reasons to think that it breaks down too.
To put it another way, the rigorous methods of calculus are derived rigorously from assumptions that are incorrect (as pertains to the real world), so they are not actually rigorous (in the context of doing physics).
In physics, an integral is a large but finite sum, and $dx$ is a small but nonzero value. Since it's nonzero, there's no reason you can't multiply and divide by it. The justification for ignoring terms of order $dx^2$ is not that they disappear in a limit, but that they're close enough to zero that the difference won't be significant amidst all of the other noise in the problem. The Dirac delta function is a function: it's a Gaussian with a width small enough that you can't see it through your slightly blurry spectacles. And so on. These approximate methods can't be proven correct, but that's okay, because in science you can never know that anything you're doing is correct.
