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I'm reviewing topics of previous calculus courses using Advanced Calculus by Taylor and Mann, and one thing that I came across that I don't understand is this part here:

Definition: Let $dx$ denote an independent variable which may take on any value whatsoever. Then the function of $x$ and $dx$ whose value is $f'(x)\,dx$ is called the differential of $f$. Observe that the differential is a homogeneous linear function of $dx$; that is, for a fixed value of $x$, the differential has as its value a fixed multiple of $dx$.

Where they treat the differentials in functions of 1 real variable as variables of the tangent line of $f(x)$ at a given point. Then they remark that the differentials can specifically take on any values, not necessarily small:

Remark: It is to be emphasized that there is no need for $dx$ and $dy$ to be small in $\frac{dy}{dx} = f'(x)$.

This somewhat collides with what I thought I knew previously, as my teachers would always refer to differentials as "infinitesimal quantities", which I had assumed meant that they should strictly take on values that tend to zero.

But then when they give a detailed explanation on what differentials are in n variables, the process by which they reach those final results seems to me like it requires that the differentials approach zero.

Here they provide the definition for functions in N variables

Though earlier than that they define what it means for a two-variable function to be differentiable at a given point, then proceed to give the definition both in terms of the quotient and in terms of the a function (A and B are the partial derivatives at those points) such that $$ df(x, y;\, dx, dy) = A dx + B dy. $$

I know also that this results from taking the limit of $$ Δf(x, y;\, \Delta x, \Delta y) = A \Delta x + B \Delta y + \varepsilon_1 Δx + \varepsilon_2 \Delta y $$ when $(\Delta x, \Delta y) \to (0,0)$ (with the $\varepsilon$ being defined so that they also tend to zero when this happens). If in order for us to get to the formula for differentials we need to make $\Delta x, \Delta y$ approach zero, then how can I also say that $dx$ and $dy$ can take on any values whatsoever?

Sammy Black
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Some random guy
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    I won't attempt to give a full answer, but I will simply point out that $dx$ has many, many different meanings in mathematics. There are a number of inequivalent ways of giving it a precise definition. – Joe Jan 15 '25 at 23:29
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    In nonstandard analysis the main property of a differential is that the product of differentials is zero. Sort of like the quadratic terms being much smaller than the linear ones in the Taylor series. so you can ignore it. – CyclotomicField Jan 15 '25 at 23:36
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    Just to add to the above comments, here's a great example of how "$dx$" works differently depending on context: quadratics in differentials don't vanish in general in these contexts. You can relate the latter to a generalization of dual numbers. – J.G. Jan 15 '25 at 23:47
  • @CyclotomicField I'm assuming by "differential" you mean an infinitesimal, and your statement is just false; the product of two nonzero infinitesimals is infinitesimal, but always nonzero. Perhaps you're thinking of smooth infinitesimal analysis, but even then (if I remember correctly) your claim is false and all we have is $\epsilon^2=0$ when $\epsilon$ is infinitesimal. – Nicholas Todoroff Jan 16 '25 at 00:33
  • This question is closely related to, but perhaps not a duplicate of, What is the difference between $dy$ and $Δy$ and why is $dx$ is same as $Δx$ when $x$ is an independent variable and $y$ is dependent one?. – Mark S. Jan 16 '25 at 13:07
  • @NicholasTodoroff is right. The general result for cross terms of differential forms is that they anticommute. – J.G. Jan 16 '25 at 15:32
  • Please note Joe's opening comment. There are many ways of defining differentials, or "infinitesimals", and those different definitions have different properties. CyclotomicField started out by stating exactly what definition he was referring to: Non-Standard Analysis. Not differential forms. And he is right: In Non-Standard Analysis, the products of infinitesimals are exactly zero. – Paul Sinclair Jan 17 '25 at 17:23
  • However, the definition of $dx$ being applied in the question is more simple-minded. $dx$ is not an infinitesimal at all. Instead $dx$ is a coordinate along the tangent line to $y=f(x)$ at a point $x$. In $dy = f'(x)dx$, $x$ controls which tangent line is being followed, and $(dx, dy)$ are points on that tangent line, with $(0,0)$ taken as the point of tangency. So $dx$ need not be small. But the tangent line is only a reasonable approximation for the function when $dx$ is small, so that is where all the important stuff occurs. – Paul Sinclair Jan 17 '25 at 17:29
  • I believe I understand now with that last comment. I still have a hard time picturing why dx and dy can attain any value even though I had to take the limit, but it seems that there are plenty of interpretations of these concepts, contrary to what I thought when I asked the original question. Thanks everybody for the help :) – Some random guy Jan 22 '25 at 01:54
  • @PaulSinclair Again, that is categorically false. In nonstandard analysis, the product of two nonzero infinitesimals is always a nonzero infinitesimal. This is a direct consequence of the Transfer Principle: the product of two nonzero real numbers is always nonzero, hence the product of two nonzero hyperreal numbers is always nonzero. It would not be a hyperreal field if it had zero divisors. Not that this is necessarily related to the question, but it bothers me that two different people have made the same blatantly false claim so confidently... – Nicholas Todoroff Jan 24 '25 at 01:52

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