1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Gold Member

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    Have you tried:

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


    User Avatar
    Science Advisor

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook