5 ms·

"dx" is a term rigorously defined in infinitesimal calculus (IMO a much easier to understand approach to differentiation.) So, yes, you can multiply by dx (even

"dx" is a term rigorously defined in infinitesimal calculus (IMO a much easier to understand approach to differentiation.) So, yes, you can multiply by dx (even if x is a more complicated function.) ∂x would have a problematic definition, as it would require selecting from its components based on information not supplied. E.g., let x = y+z. Then dx = dy + dz. But ∂x (by extension) = ∂y + ∂z, but at least one of these terms on the right is identically zero, depending on whether y or z is held constant. So ∂x doesn't have a meaning.

Koshkin 4y ago
Actually, “dx” is rigorously defined in the standard calculus, too, and yes, you can multiply (and divide) by it.
- oolveea 4y ago
  For the love of anything sacred, can you please point me to a resource that explains all possible operations with “dx” and their conceptual meaning. Like, I get “dx”, but I cannot put my finger to it!!! This might be because the Precise Definition of the Limit phrases it as “x approaches a”; it is as though we are “sent” to the land of dx, but not told what it is as an atomic concept!
  - bmitc 4y ago
    I am not entirely sure what the above commenter means that dx is rigorously defined in "infinitesimal calculus" because I don't know what they necessarily mean by "infinitesimal calculus". As far as I am aware, there is standard calculus, non-standard analysis by Robinson, and smooth infinitesimal analysis that uses intuitionistic logic. The three are very different. dx has no meaning in standard calculus. It is simply there for notation. It is given meaning by the theory of smooth manifolds and differential forms. In that setting, differentials such as dx are given explicit meaning: they are functions that operate on tangent vectors. For example, apply dx to the unit vector d/dx + d/dy to get dx(d/dx+ d/dy) = d/dx(x) + d/dy(x) = 1 + 0 = 1.
    - Koshkin 4y ago
      > dx has no meaning in standard calculus Sure it does. There is no need to know about smooth manifolds or differential forms to understand the differential of a function of one variable at a point and the meaning of dy = f’(x)dx.
      - oolveea 4y ago
        Can you please point me to a resource that explains all possible operations with “dx” and their conceptual meaning.
        Koshkin 4y ago
        Any introductory calculus book worth the paper it’s printed on would gladly tell you that the differential of the function y = x at a point x0 is nothing more than x - x0 and that you do not have to think about it as something that is “infinitely small” or anything equally mysterious. (Some would even go as far as saying that “the differential of a function of one variable is a linear map of the increment of the argument.”) So, with dx = x - x0, you can do with it anything you want, even divide by it (assuming that dx stays non-zero).
      - bmitc 4y ago
        What is the meaning then? dy = f'(x)dx is just a definition for notional convenience, primarily employed when doing u or u-v substitution. My point is that dx in single variable calculus is notation. It is not an intrinsic object. dx is an intrinsic object as a differential form on a smooth manifold. Of course, the real line R is a 1-manifold, so dx does have that meaning, but you need to understand what a differential form is to know that. One doesn't necessarily need the full generality of smooth manifolds though. Harold Edwards' Advanced Calculus: A Differential Forms Approach and Advanced Calculus: A Geometric View teach differential forms for Euclidean manifolds.
        bmitc 4y ago
        The second book is by James Callahan. I accidentally left that off.
  - mjthrowaway1 4y ago
    Thank you for asking this. This plagued me for years as an undergrad physics student.
Blammar 4y ago
This is one book on infinitesimal calculus: https://www.amazon.com/Infinitesimal-Calculus-Dover-Books-Mathematics/dp/0486428869 https://www.amazon.com/Infinitesimal-Calculus-Dover-Books-Ma... . However, and this is very amusing to me, it turns out that the process of automatic differentiation (see https://en.wikipedia.org/wiki/Automatic_differentiation https://en.wikipedia.org/wiki/Automatic_differentiation, the section on dual numbers) works in exactly the same way. Just replace all of the primed symbols (u', v', etc.) with du, dv, etc. and dual numbers are isomorphic to infinitesimals (if I'm using that term correctly.)

Koshkin 4y ago

Actually, “dx” is rigorously defined in the standard calculus, too, and yes, you can multiply (and divide) by it.

oolveea 4y ago
For the love of anything sacred, can you please point me to a resource that explains all possible operations with “dx” and their conceptual meaning. Like, I get “dx”, but I cannot put my finger to it!!! This might be because the Precise Definition of the Limit phrases it as “x approaches a”; it is as though we are “sent” to the land of dx, but not told what it is as an atomic concept!
- bmitc 4y ago
  I am not entirely sure what the above commenter means that dx is rigorously defined in "infinitesimal calculus" because I don't know what they necessarily mean by "infinitesimal calculus". As far as I am aware, there is standard calculus, non-standard analysis by Robinson, and smooth infinitesimal analysis that uses intuitionistic logic. The three are very different. dx has no meaning in standard calculus. It is simply there for notation. It is given meaning by the theory of smooth manifolds and differential forms. In that setting, differentials such as dx are given explicit meaning: they are functions that operate on tangent vectors. For example, apply dx to the unit vector d/dx + d/dy to get dx(d/dx+ d/dy) = d/dx(x) + d/dy(x) = 1 + 0 = 1.
  - Koshkin 4y ago
    > dx has no meaning in standard calculus Sure it does. There is no need to know about smooth manifolds or differential forms to understand the differential of a function of one variable at a point and the meaning of dy = f’(x)dx.
    - oolveea 4y ago
      Can you please point me to a resource that explains all possible operations with “dx” and their conceptual meaning.
      - Koshkin 4y ago
        Any introductory calculus book worth the paper it’s printed on would gladly tell you that the differential of the function y = x at a point x0 is nothing more than x - x0 and that you do not have to think about it as something that is “infinitely small” or anything equally mysterious. (Some would even go as far as saying that “the differential of a function of one variable is a linear map of the increment of the argument.”) So, with dx = x - x0, you can do with it anything you want, even divide by it (assuming that dx stays non-zero).
    - bmitc 4y ago
      What is the meaning then? dy = f'(x)dx is just a definition for notional convenience, primarily employed when doing u or u-v substitution. My point is that dx in single variable calculus is notation. It is not an intrinsic object. dx is an intrinsic object as a differential form on a smooth manifold. Of course, the real line R is a 1-manifold, so dx does have that meaning, but you need to understand what a differential form is to know that. One doesn't necessarily need the full generality of smooth manifolds though. Harold Edwards' Advanced Calculus: A Differential Forms Approach and Advanced Calculus: A Geometric View teach differential forms for Euclidean manifolds.
      - bmitc 4y ago
        The second book is by James Callahan. I accidentally left that off.
- mjthrowaway1 4y ago
  Thank you for asking this. This plagued me for years as an undergrad physics student.

Blammar 4y ago

This is one book on infinitesimal calculus: https://www.amazon.com/Infinitesimal-Calculus-Dover-Books-Mathematics/dp/0486428869 https://www.amazon.com/Infinitesimal-Calculus-Dover-Books-Ma... . However, and this is very amusing to me, it turns out that the process of automatic differentiation (see https://en.wikipedia.org/wiki/Automatic_differentiation https://en.wikipedia.org/wiki/Automatic_differentiation, the section on dual numbers) works in exactly the same way. Just replace all of the primed symbols (u', v', etc.) with du, dv, etc. and dual numbers are isomorphic to infinitesimals (if I'm using that term correctly.)