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Correct terminology for taking differentials?

  1. Dec 7, 2014 #1

    bcrowell

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    Consider the following two calculations:

    (1) [itex]d(x\cos x)=(\cos x-x\sin x)dx[/itex]

    (2) [itex]\frac{d}{dx}(x\cos x)=(\cos x-x\sin x)[/itex]

    I would describe these both as differentiation. Is there a standard terminology that allows one to make the distinction between the two, if desired? The best I could come up with was "taking a differential" for 1, as distinguished from "differentiating a function" for 2. But these are both awkward, and I don't know if they're standard or would be widely understood.
     
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  3. Dec 7, 2014 #2

    dextercioby

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    (2) is the <differentiation of a 1-var function> which basically is a mapping from a set of functions to a set of functions, that is an operator (linear mapping) which takes a function and 'spits' a function. To the 1st formula, I see a nice interpretation in terms of differential geometry: applying the exterior differential on a scalar (0-form field) gives a 1-form field.
     
    Last edited: Dec 7, 2014
  4. Dec 7, 2014 #3

    bcrowell

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    Thanks for your reply, but I wasn't asking for an interpretation. I was asking about terminology. (There are many different interpretations of the Leibniz notation, going back over 300 years. All of these interpretations are equally valid.)
     
  5. Dec 7, 2014 #4

    dextercioby

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  6. Dec 7, 2014 #5

    bcrowell

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    OK. I was asking for terminology for the operation or process. So would you say that the terms I suggested originally are the best available, and/or are widely understood?
     
  7. Dec 7, 2014 #6

    dextercioby

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    By "differentiating f(x)" everyone understands "calculating the 1st derivative of f(x) (provided it exists) at a generic point x of the function's domain". That's your (2). "taking/computing/considering the 1st differential of f(x)" would naturally depict your 1st option. There's not too much room for terminology variation here, really.

    If (indefinite) integration (of a function) is an operation, then "taking a differential (of a function)" is the reverse operation (up to a so-called 'integration constant') . ## \int d f = f +C##

    Perhaps someone else can better address your question. I only wrote what I knew. :)
     
    Last edited: Dec 7, 2014
  8. Dec 7, 2014 #7

    bcrowell

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    OK, thanks for your help -- much appreciated!
     
  9. Dec 8, 2014 #8
    If someone asked me to differentiate [tex]x^2cos(x)[/tex] (with respect to x implied because it's the only variable in the expression), I would do what you did in (2).

    So, differentiation = the act or process of taking a derivative.

    Now, whether "finding the differential" can be referred to as differentiation or not depends on what school of thought you are following.

    Following the modern/"standard" interpretation, the act of finding a differential isn't really differentiation, although differentiation is a step you need to perform. This is because a differential is defined in terms of the derivative. (More specifically, it is defined as a change in the linear approximation of a function, which automatically brings in the derivative.)

    In contrast, a proponent of Nonstandard Calculus may argue that finding the differential IS differentiating followed by a minor rearrangement. In other words, he would be interpreting the differential as an infinitesimal change in the function rather than as a change in the linear approximation. In that sense, all you are doing in (1) is differentiating.

    To be safe, I would only refer to (2) as differentiating and refer to (1) as either "finding the differential" or "calculating the differential." I don't see anything wrong with "taking the differential" either, although, as you said, it does sound a bit awkward.
     
  10. Dec 9, 2014 #9
    The second method is finding the derivative of a function, i.e. it's rate of change at a point, the first method is finding the exterior derivative of a function (i.e. a 0-form) which is a method of finding the field of sheets (i.e. density) generated by this form
    http://math.stackexchange.com/quest...on-zero-closed-loop-line-integrals-via-sheets

    In other words, the second method takes the perspective of functions as curves, surfaces etc... the other views functions as densities, and we geometrically represent these densities by sheets, how many sheets (intersecting sheets, e.g. 2 forms etc...) do you mass through as you go along a curve/surface etc...

    Thus the terminology for the first is 'taking an exterior derivative' and for the second your 'taking a derivative'.
     
    Last edited: Dec 9, 2014
  11. Dec 9, 2014 #10

    PeroK

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    I'd say the first is "differentials" and the second is "differentiation" or the "derivative". Usually, one would define the derivative first:

    ##\frac{dy}{dx} = \lim_{h \rightarrow 0} \frac{y(x+h)-y(h)}{h}##

    Then, define the differential as:

    ##dy = \frac{dy}{dx} dx##

    This would be standard if developing calculus rigorously (in my experience, anyway).
     
  12. Dec 13, 2014 #11
    It doesn't seem to me that anyone with any reasonable experience could possibly misinterpret the terminology you used in your original post.

    Chet
     
  13. Dec 13, 2014 #12

    atyy

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    (1) exterior differentiation
    (2) differentiation
     
  14. Dec 14, 2014 #13

    Stephen Tashi

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    Discussions of differentials on the forum usually don't have a clear resolution! There are various opinions.

    My opinion:

    If you are talking about differential geometry then differentials like [itex] dx [/itex] have a well defined interpretation (which I'd have to look up). However, on the forum, most posts using differentials aren't talking about differential geometry.

    You number 2. is differentiation.

    Your number 1. is differentiation. It's differentiation using abiguity. People applying calculus tend to use symbols like [itex] x [/itex] ambiguously. They use them to stand for both variables and functions. If [itex] x [/itex] is a function of some other variable [itex] t [/itex] then you can treat [itex] dx [/itex] as a reminder to include the factor [itex] x'(t) = \frac{dx}{dt} [/itex] in the derivative.
     
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