4 ms·

But you do have 1/(dy/dx) = dx/dy and you can get the insight deep if you follow on with actually using differentials properly, as in the paper linked here in t

But you do have 1/(dy/dx) = dx/dy and you can get the insight deep if you follow on with actually using differentials properly, as in the paper linked here in this very thread by LudwigNagasena. But then of course, you see that d^2 f/dx^2 is a bad expression for the second derivative, and that you should actually use differentials properly and write it as (d^2 f)/(dx)^2 - df/dx (d^2 f)/(df)^2.

bikenaga 3y ago
The second derivative notation makes sense when you think of d/dx as an operator on functions. (As someone noted below, d/dx is essentially a tangent vector, which is something that operates on functions.) d/dx takes the derivative of what follows with respect to x, so (d/dx)(f) = df/dx is the derivative function of f with respect to x. Now you want to differentiate df/dx with respect to x, so you do "d/dx" to "df/dx": (d/dx) (df/dx) (This looks better if you write it with "real" fractions.) On the top you have "d d f" with is naturally written "d^2f". On the bottom, you have "dx dx", which is naturally written "dx^2". So the result should be written d^2 f/dx^2.
- impossiblefork 3y ago
  The problem is that d/dx(dy/dx) is not d^2 y/(dx)^2. d/dx (dy/dx) is really d(dy/dx)/dx, which is ((d^2 y)/dx + dy d(1/dx))/dx = d^2 y/dx - dy/dx (d^2 x)/(dx)^2, where df is the actual differential.
  - bikenaga 3y ago
    We're probably talking about different things. I'm not using "d/dx" as the differential operator "d" divided by the differential "dx", which is what I think you mean. I mean the compound symbol "d/dx" from calculus, which means differentiation with respect to x - or alternatively (from a differential geometric point of view) the tangent vector "d/dx" operating on the function f. I meant to explain why the notation "d^2 f/dx^2" is the way it is. It looks funny, and students would often ask why the 2's were "in different places". But it makes sense when you remember that "d/dx" is the derivative operator.
    - impossiblefork 3y ago
      I am talking about d/dx as the differential operator d divided by dx. That's the way to make it actually work, algebraically. d^2 f/dx^2 is actually wrong as the way to write the second derivative unless df/dx = 0 or d^2 x/(dx)^2 = 0; the first case is trivial, and the second is almost never the case. Suppose for example that in actually x = t^2. Then dx = 2t dt, d(dx) = 2(dt)^2 + 2 t (d^2 t) and we get (2(dt)^2 + 2t(d^2 t))/(2t dt)^2 = 0.5 + (...) d^2 t/(dt)^2, so it certainly isn't zero. But if you have actual differentials you can do algebra.
      - bikenaga 3y ago
        > I am talking about d/dx as the differential operator d divided by dx. Okay, that's what I figured from your earlier computation. (One note: If "d" means exterior differentiation, then d^2 = 0.) But if you mean d/dx in this way you should make sure you tell people, because that's not the interpretation it has in calculus (say a typical calc class in a typical college) or in differential geometry. In those cases, d/dx means (calc class) the differentiation operator, or a tangent vector (diff geom) [which is again an operator on functions]. That is, "d/dx" is regarded as a single symbol, not a quotient of exterior derivative by a differential. > d^2 f/dx^2 is actually wrong as the way to write the second derivative ... I think this is true given your interpretation, based on your earlier computation. But again, in ordinary calc classes (which after all don't talk about differential forms) the "d^2 y/dx^2" notation comes from the standard calc class meaning for d/dx and it goes back ages. I just looked in G. H. Hardy's classic "Pure Mathematics" from over a hundred years ago and yep! - d^2 y /dx^2 is one of the notations for the second derivative. (I'm not sure if it goes back to Leibniz.) So if you're lobbying for a change you have your work cut out for you. :-) Regardless, I hadn't thought of interpreting "d/dx" as "exterior derivative divided by differential" - interesting idea. [edited: added note about d^2, minor edit in last sentence]
        impossiblefork 3y ago
        It's clearly not exactly an exterior derivative. d(df) can't be seen as zero, since we will be dividing it by dx, and we will therefore need to retain things that would otherwise become infinitesimal.

bikenaga 3y ago

The second derivative notation makes sense when you think of d/dx as an operator on functions. (As someone noted below, d/dx is essentially a tangent vector, which is something that operates on functions.) d/dx takes the derivative of what follows with respect to x, so (d/dx)(f) = df/dx is the derivative function of f with respect to x. Now you want to differentiate df/dx with respect to x, so you do "d/dx" to "df/dx": (d/dx) (df/dx) (This looks better if you write it with "real" fractions.) On the top you have "d d f" with is naturally written "d^2f". On the bottom, you have "dx dx", which is naturally written "dx^2". So the result should be written d^2 f/dx^2.

impossiblefork 3y ago
The problem is that d/dx(dy/dx) is not d^2 y/(dx)^2. d/dx (dy/dx) is really d(dy/dx)/dx, which is ((d^2 y)/dx + dy d(1/dx))/dx = d^2 y/dx - dy/dx (d^2 x)/(dx)^2, where df is the actual differential.
- bikenaga 3y ago
  We're probably talking about different things. I'm not using "d/dx" as the differential operator "d" divided by the differential "dx", which is what I think you mean. I mean the compound symbol "d/dx" from calculus, which means differentiation with respect to x - or alternatively (from a differential geometric point of view) the tangent vector "d/dx" operating on the function f. I meant to explain why the notation "d^2 f/dx^2" is the way it is. It looks funny, and students would often ask why the 2's were "in different places". But it makes sense when you remember that "d/dx" is the derivative operator.
  - impossiblefork 3y ago
    I am talking about d/dx as the differential operator d divided by dx. That's the way to make it actually work, algebraically. d^2 f/dx^2 is actually wrong as the way to write the second derivative unless df/dx = 0 or d^2 x/(dx)^2 = 0; the first case is trivial, and the second is almost never the case. Suppose for example that in actually x = t^2. Then dx = 2t dt, d(dx) = 2(dt)^2 + 2 t (d^2 t) and we get (2(dt)^2 + 2t(d^2 t))/(2t dt)^2 = 0.5 + (...) d^2 t/(dt)^2, so it certainly isn't zero. But if you have actual differentials you can do algebra.
    - bikenaga 3y ago
      > I am talking about d/dx as the differential operator d divided by dx. Okay, that's what I figured from your earlier computation. (One note: If "d" means exterior differentiation, then d^2 = 0.) But if you mean d/dx in this way you should make sure you tell people, because that's not the interpretation it has in calculus (say a typical calc class in a typical college) or in differential geometry. In those cases, d/dx means (calc class) the differentiation operator, or a tangent vector (diff geom) [which is again an operator on functions]. That is, "d/dx" is regarded as a single symbol, not a quotient of exterior derivative by a differential. > d^2 f/dx^2 is actually wrong as the way to write the second derivative ... I think this is true given your interpretation, based on your earlier computation. But again, in ordinary calc classes (which after all don't talk about differential forms) the "d^2 y/dx^2" notation comes from the standard calc class meaning for d/dx and it goes back ages. I just looked in G. H. Hardy's classic "Pure Mathematics" from over a hundred years ago and yep! - d^2 y /dx^2 is one of the notations for the second derivative. (I'm not sure if it goes back to Leibniz.) So if you're lobbying for a change you have your work cut out for you. :-) Regardless, I hadn't thought of interpreting "d/dx" as "exterior derivative divided by differential" - interesting idea. [edited: added note about d^2, minor edit in last sentence]
      - impossiblefork 3y ago
        It's clearly not exactly an exterior derivative. d(df) can't be seen as zero, since we will be dividing it by dx, and we will therefore need to retain things that would otherwise become infinitesimal.