# Why do we treat differentials as infinitesimals, even when it's not rigorous

From single-variable calculus where we first encounter differentials we are told fairly often that differentials are not to be treated as infinitesimal quantities/objects (but we are never really told why), and we kind of hand-wave things and manipulate them as fractions (e.g $\frac{dy}{dx}$) in things like $u$-Substitution.

$$\int f(x) \ dx$$

In the above integral, the differential, $dx$, seems to be doing absolutely nothing other than signifying where the expression for the integrand finishes. But the differential, seems to have more of a part to play in multivariable calculus.

In multivariable calculus we often encounter integrals like

$$\oint \vec{f(x)}\bullet\vec{dX} = \oint \|\vec{f(x)}\| \cdot \|\vec{dX}\| \cos(\theta)$$

where we can treat vector differentials as if they were infinitesimal vectors, and perform normal vector operations (such as taking the dot product)

• It doesn't seem all that clear to me why we can't treat either scalar or vector differentials as infinitesimals.
• Furthermore why do we even treat differentials as infinitesimals, even when it's not technically rigorous?
• I'm sure there must be conditions that need to be met that allow us to treat differentials as infinitesimals.

It seems like in single-variable calculus the differentials we encountered just a special case of the differentials we encounter in multivariable calculus, which we can do more with.

I've heard of integration on differential forms (which I'm about to start learning), and I'm assuming that all the differentials we've encountered thus far in single and multivariable calculus must be special cases of this general form. Am I correct? If not then where do we rigorously define differentials and the operations we can perform with them?

• – dxiv Sep 27 '16 at 6:31
• It is a substantial mistake to treat current understanding of rigor as the last word. Logic will continue to advance. – Michael Hardy Sep 27 '16 at 7:11
• @Perturbative : No --- that's the opposite of what I had in mind. I meant mathematicians' understanding generally. $\qquad$ – Michael Hardy Sep 27 '16 at 7:29
• Your bounty offer is basically asking for someone to write a survey paper for you. That's excessive for an SE post, in my opinion; questions like that tend to get closed as "too broad". – Nate Eldredge Oct 1 '16 at 2:58
• We don't treat differentials as infinitesimals in introductory calculus because introductory calculus students don't know how to work with the various varieties of infinitesimals either. In standard analysis, differentials and related notions are way we give the notion of infinitesimal meaning. The two most common first places to see differential forms defined rigorously, I think, are introduction to differential geometry and introduction to algebraic geometry. – user14972 Oct 1 '16 at 4:16

Treating a differential as an infinitesimal isn't always guaranteed to make sense. For instance if I'm integrating over some smooth manifold other than $\mathbb{R}$, the concept of an infinitesimal might not be well-defined. In essence, differential forms were constructed in a particular way to answer some of your questions.
Given some smooth manifold $X$, a differential form $d\omega$ is a function $X \to \Lambda(T_{x}(X)^{*})$. That is to say at each point $x$, $d\omega(x)$ gives us an alternating tensor on the tangent space at $x$. If you squint hard enough, you'll see that this amounts to measuring how much the derivative changes at a point, somewhat similar to the familiar $dx$ in $\mathbb{R}$.
Now differential forms are also anticommunative, $dx \wedge dy = -dy \wedge dx$, which implies $dx \wedge dx = 0$.
So it's not fair to treat $dx$ as an infintesimal as that won't always make sense, and it doesn't generalize well. This kind of went off on a tangent, but differential forms are pretty cool and if you're interested I'd recommend the last chapter of Guillemin and Pollack (and if you're rusty on the prerequisites, chapters 1 and 2 of the same book). I'd say it's worth the price of admission to prove the generalized Stokes theorem, which is in my opinion on of the most elegant theorems around.