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Meaning of Dy/Dx

  1. Oct 2, 2012 #1
    Hello all~

    I really was curious as to why dy/dx can be treated as a fraction when solving equations, and not an operator. For example why can

    dy/dx = x be the same as dy = xdx?

    My second question was what EXACTLY does dx (any variable, I'll just call it x) mean by itself?

    Finally, why is indefinite integral notation written (integral sign) f(x) dx? What is the significance of the dx?

  2. jcsd
  3. Oct 2, 2012 #2
    dy/dx should not be treated as a fraction if you are talking about the leibniz notation for differentiation.

    dx basically represents change in x in an integral. It tells you that is the variable that is being integrated.
  4. Oct 2, 2012 #3
    When is dy/dx allowed to be treated as a fraction apposed to an operator?

    What properties of the integral symbol allows a function (y prime) to be transformed back to f(x)?
  5. Oct 2, 2012 #4
    Why does the anti-derivative take the form

    (integral symbol) f'(x) dx?
  6. Oct 3, 2012 #5
    If dy/dx is the leibniz notation then it is always an operator. It should never be treated as a fraction.

    For the integral symbol, I think that's just what it is.. a symbol.
  7. Oct 3, 2012 #6


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    I have said this a number of times on this forum: dy/dx is NOT a fraction but it can be "treated" like a fraction. That is because it is defined as the limit of a fraction. So we cannot prove the chain rule, dy/dx= (dy/du)(du/dx), just by saying "cancel the 'du's" but we can go back before the limit in its definition, cancel there, then take the limit (that is the standard proof of the chain rule in Calculus texts).

    So if we have something really complicated like "(dy/dv)(dz/dt)(dv/dz)" it doesn't hurt to think "Aha! I can 'cancel' the "dv"s and "dz"s to get dy/dt" as long as you understand you are NOT actually canceling, you are using the chain rule.

    Again, derivatives are NOT fractions but they can be "treated" like fractions.

    Historically, the "[itex]\int[/itex]" symbol is from "S", for "sum", referring to the Riemann sums whose limit gives the integral. (My father always called it a "seahorse"!)
  8. Oct 3, 2012 #7


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    No. The operator is ##\frac{d}{dx}##. dy/dx represents the derivative of y with respect to x. The operator d/dx is operating on y.
  9. Oct 3, 2012 #8


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    This is too strong, because the normal definition of "fraction" is the quotient of two real numbers, but you can run into contradictions if you treat dy/dx like a fraction and assume that all the axioms of the real-number system hold.

    A more accurate way to say this is that dx and dy are members of an extended number system that augments the real numbers with infinitesimally small numbers. This number system (known as the hyperreal numbers) obeys all the elementary axioms of the real number system ( http://en.wikipedia.org/wiki/Real_number#Axiomatic_approach ) , where "elementary" means everything but the completeness property. Since most freshman calc students have never heard of the completeness property, they will be in no danger of using it inadvertently. You also have to interpret the derivative of y with respect to x not as the quotient dy/dx but as the "standard part" of that quotient, meaning the real number nearest to it.

    Nice discussion here: http://math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio
  10. Oct 3, 2012 #9
    So, basically what you're saying is that a true fraction can't contain hyperreal quantities (in this case infinitesimals)? Or am I just confused?
  11. Oct 3, 2012 #10
    Someone in my calculus class asked this one day, and my professor responded by saying that [itex]\frac{dy}{dx}[/itex] is not a fraction and should not be treated as a fraction; it is treated as one when solving a differential equation by separation of variables, but he said that that was just because it happened to work. Also, there are apparently two different notations of expressing derivatives: one of them treats [itex]\frac{dy}{dx}[/itex] as a fraction, while the other does not, though I don't know much more about this. There is also a way of thinking of dy, or other variable, as an infinitesimal number, which I don't know much about either.

    For your second question, to my knowledge, the d in front of dx or another variable is representative of delta and is symbolic of the change of x, or whatever other variable it is. Therefore [itex]\frac{dy}{dx}[/itex] is the change in the y value with respect to the x value; this is related to the fact that the derivative gives you the instantaneous slope at a given point.

    For your final question, the dx represents the change in the x value. When you take the integral of a function, it gives you the area below the graph. This is because the height of a function is represented by f(x), which is getting multiplied by dx, which represents an infinitely small width.

    You can go a lot further in-depth with this by looking at the limit definitions of an integral and derivative, and seeing how they work together, though it can get quite complicated. I gained a lot of feeling for what an integral and derivative are due to seeing these limit definitions and thinking about how they relate to each other.
  12. Oct 3, 2012 #11


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    Your prof is oversimplifying (and/or you're misquoting him and/or he's mistaken and/or his training is 50 years or more out of date). For a correct answer, see the link at #8.
  13. Oct 4, 2012 #12
    Using hyperreals is very dangerous, because most textbooks and mathematicians do not work with these objects. The standard way of dealing with [itex]\frac{dy}{dx}[/itex] is that it is not a fraction but can be treated as such.
    If you wish to understand why [itex]\frac{dy}{dx}[/itex] can be treated as a fraction, then you can find an answer in nonstandard analysis where it is shown that it actually can be interpreted as a fraction of hyperreal numbers. But it is important to know that the vast majority of people do not work with infinitesimals. So interpreting [itex]\frac{dy}{dx}[/itex] as a fraction is alright, as long as you know it is not the mathematical convention.
  14. Oct 4, 2012 #13
    Gosh haven't we got it all mixed up; no wonder the OP (and others) are confused.

    There are several different balls in play here.

    1) First consider the function y = f(x)

    There is another function (of x) called the derived function or derivative.

    This is a valid function that you can plot (draw the graph of) and calculate values of for different values of x (in the range of interest.)

    We call it variously f' or rather loosely dy/dy. edit: dy/dx or better d/dx

    In this sense dy/dx can be considered as an operator operating on the function f(x) to yield another function.

    Sometimes we use the capital D for this (the D operator).

    2) We can consider it the ratio infinitesimals, small quantities and take limits etc.

    This yields the slope of the curve y = f(x) at a particular point and is a pure number.

    It is part of above-high school courses in calculus/analysis to show that these two approaches are equivalent.

    You should also be aware that capital D is also used in fluid mechanics for a slightly different form of differentiation called differentiation following the flow (Eulers method) to distinguish it form differentiation in relation to fixed axes.
    Last edited: Oct 4, 2012
  15. Oct 4, 2012 #14


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    Perhaps you are the one who is "mixed up". While true, nothing you say is relevant to the original question.
  16. Oct 4, 2012 #15
    Well in fairness, the original question uses a capital Dy/Dx

    The only place I have seen this is in fluids as stated and I am the only one to note this so, strictly speaking, I am the only respondent to answer the actual question asked.

  17. Oct 4, 2012 #16


    Staff: Mentor

    The operator is ##\frac{d}{dx}## not ##\frac{dy}{dx}##. If I put what you wrote into symbols, I would have ## \frac{dy}{dx}(f(x))##, which would be interpreted as the product of a derivative and f(x) rather than the derivative of f(x).
    Assuming D means differentiation w. respect to x, then D = ##\frac{d}{dx} ##.
  18. Oct 4, 2012 #17
    Mark, yes that's perfectly true, but I did say 'rather loosely'.

    Thank you for correcting me perhaps I should say

    We call it variously f' or rather loosely d/dx.

    I note incidentally a typo here that I have edited.

    I suspect that the OP is being taught a particular approach, but has seen other apparently conflicting statements when looking elsewhere and is confused, since he has subsequently posted another question along similar lines in the differential equations forum.

    I am trying not to upset the line of teaching undertaken but to support it.
  19. Oct 4, 2012 #18
  20. Oct 4, 2012 #19


    Staff: Mentor

    Maybe, extremely loosely. f' is the function that represents the derivative of f. A better pairing, in my view, would be f' and dy/dx.

    If you want to talk about the value of the derivative at a particular number x0, we have f'(x0) and $$ \left. \frac{dy}{dx}\right |_{x = x_0}$$

    My point is that we can evaluate f' at an arbitrary number x using the notation f'(x), but the notation below is meaningless:
    $$ \left. \frac{d}{dx}\right |_{x}$$
  21. Oct 4, 2012 #20
    Ah I remember asking this question before on this forum.
    Some more information is available on this thread which I made a year or so ago, about the manipulation of differentials in differential equations:


    It might answer your question.

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