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Dy/dx - Fraction and/or Operator

  1. Sep 27, 2011 #1
    I am a bit confused over the use of the derivative operator dy/dx. I realise dy is a very small change in y and dx is a very small change in x. When combined into dy/dx it is an operator which means take the derivative of y with respect to x.
    However I notice many authors still treat it as a fraction- a small change in y over a small change in x. eg
    di=[itex]\frac{1}{L}[/itex]v dt
    [itex]\int[/itex]di=[itex]\frac{1}{L}[/itex][itex]\int[/itex]v(t) dt

    Everything works out nicely but it is a bit confusing when operators can be treated as fractions.
  2. jcsd
  3. Sep 27, 2011 #2

    Char. Limit

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    Strictly speaking, dy/dx cannot be taken as a fraction. However, as an abuse of notation in SOME cases, they will use this as a fraction regardless. Usually, this can be done rigorously another way. For example, if you'll allow multiplication by differentials...

    [tex]v(t) = L \frac{di}{dt}[/tex]

    [tex]\frac{1}{L} v(t) dt = \frac{di}{dt} dt[/tex]

    [tex]\int \frac{1}{L} v(t) dt = \int \frac{di}{dt} dt[/tex]

    Then using the u-substitution on the right-hand side, di/dt dt simplifies to di, and we get the problem as originally stated.

    [tex]\frac{1}{L} \int v(t) dt = \int di[/tex]
  4. Sep 27, 2011 #3
    I have found a bit more clarity thinking of dy/dx as a _differential_ operator, which assigns to a differentiable f its differential f'(t)dt , which is the local-linear approximation to the change of values of f, but there may be some uses (and maybe abuses) of notation that I am not familiar with.
  5. Sep 28, 2011 #4


    Staff: Mentor

    The derivative operator is d/dx, not dy/dx. The first symbol operates on a differentiable function of x. The second symbol represents the derivative (with respect to an independent variable x) of a differentiable function y.

    The differential operator is usually written as d, as in d(t2) = 2t dt.
  6. Sep 28, 2011 #5
    You're right, Mark44 , d/dt is the usual format for the operator assigning the
    differential . Moreover, the differential of a differentiable function is a differential form.
  7. Sep 30, 2011 #6
    I think I have figured it out?
  8. Oct 3, 2011 #7
    Looks right. The resulting derivative, at least.
  9. Oct 4, 2011 #8


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    As Mark44 said, d/dx is the "operator", not dy/dx.

    dy/dx is NOT a fraction- but it can be treated like one. Specifically, dy/dx is the limit of the "difference quotient" (f(x+h)- f(x))/h. So you can "go back before the limit", use the appropriate fraction property, and then take the limit.

    To make that "treat the derivative as a fraction" rigorous, we define the "differentials" dx and dy separately- though most elementary texts just "hand wave" those definitions.
    Last edited by a moderator: Mar 1, 2012
  10. Oct 4, 2011 #9
    Well, if y(x) is differentiable, then dy is f'(x)dx, and dx is just dx.

    dy is the change of y(x) along the tangent line (seen as a limiting position of the secant).

    But I agree that a lot of texts on PDE's just happily cross-multiply in cases of separation of

    variables, without justification.
  11. Feb 29, 2012 #10
    I am revisiting ODE's, and this doubt is killing me as well.

    I am re-learning by ( ODE's ( Tenenbaum and Pollard) from Dover.

    It is pretty clear that dy/dx represents f`(x), as it is also very easy to understand "geometrically" that dy = f`(x) dx.
    However, altough I am confortable solving ODE's, I just don't understant how the hell is it mathematically possible to multiply an equation by dx in order to integrate separately.
    For example:

    Q(x,y) dy/dx + P(x,y) = 0.
    P(x,y) dx + Q(x,y) dy = 0.

    They treat dy/dx as a fraction. I would like to know why, how is that possible.
    IN fact I would love if you guys could recommend me a good book that explains this very clearly.
  12. Feb 29, 2012 #11
    I like HallsofIvy's explanation. Its a couple of posts above
  13. Feb 29, 2012 #12


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    Have you tried:

    http://en.wikipedia.org/wiki/Separation_of_variables ?
  14. Apr 12, 2012 #13


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