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Dividing out differentials

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi

    Say I have the equality
    [tex]
    \frac{df(x)}{dx} = \frac{dg(z)}{dz}
    [/tex]
    where f and g are two functions that are well-behaved such that I can take their derivate. The variables x and z are both real, and run from -∞ to ∞. In this case, am I allowed to divide out the differentials dx and dz such that
    [tex]
    df(x) = dg(z)
    [/tex]
    which I can integrate and obtain
    [tex]
    f(x) = g(z) + \text{constant}
    [/tex]
    ?

    Best,
    Niles.
     
  2. jcsd
  3. Sep 9, 2012 #2
    I'm assuming from here on that f and g are functions of one variable...

    You can come to this conclusion without doing any formal calculations (eg. "dividing out by differentials"). Although what you have done would work.

    z is just a variable, so you can just write g(x) instead of g(z). Then, you get
    [tex]
    \frac{df(x)}{dx}=\frac{dg(x)}{dx}.[/tex] Or, more sggestively, [itex]f'(x)=g'(x)[/itex]. Then, integrating both sides, you get [itex]f(x)=g(x)+constant[/itex].

    Dividing out by differentials, which is a formal calculation (and thus not rigorous), is not necessary. Although, it is true that, in a sense, you can divide them out (see differential 1-forms).
     
    Last edited: Sep 9, 2012
  4. Sep 9, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Once we have the derivative, dy/dx, it is standard to define the "differential" by [tex]dy= (dy/dx)dx[/tex] where we can think of the "dx" as purely "symbolic". While the derivative is NOT a fraction, it is the limit of a fraction, the "difference quotient". It is possible to prove any "fraction property" by going back before the limit, applying the property to the difference quotient, then taking the limit. To make use of the fact that the derivative can be treated like a fraction, we define "differentials" in that way- we can say
    [tex]f'(x)= \frac{dy}{dx}[/tex]
    where the f'(x) on the left is the derivative and the dy/dx on the right is the ratio of the differentials.
     
  5. Sep 9, 2012 #4
    Thanks for the quick replies.
     
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