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Differentiation Variable

  1. May 2, 2009 #1
    I'm working out some problems, and I'm ending up with a term similar to the following:

    du/d(y/u)

    I'm differentiating with respect to y/u. Both y and u are variables. How can I divide that up to represent differentiation with just one variable (Even if it means expanding the term)?

    Is it mathematically possible to do that?

    Thanks.
     
  2. jcsd
  3. May 2, 2009 #2
    I'm really sure what you're trying to do, but if you have a function [tex]f(x,y)[/tex] and you want to know how [tex]f[/tex] changes in the limiting case with respect to a change in the ratio of [tex]y/x[/tex], then this is a case for the directional derivative. Note that [tex]v = (x, y)[/tex] is the vector where [tex]y/x[/tex] stays constant, so you'd want to take the vector orthogonal to that; ie, use [tex]v = (y, -x)[/tex]. In that case, [tex]\nabla_v f(x,y) = \nabla f(x,y) \cdot v[/tex]
     
  4. May 3, 2009 #3

    HallsofIvy

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    Let v= u/y. Then, if f(u,y) is any function of u and y, [itex]df/dv= \partial f/\partial u \partial u\partial v+ \partial f/\partial y \partial y/partial v[itex].

    Since, here, v= u/y, so u= yv and [itex]\partial u/\partial v= y[itex]. Similarly, y= u/v so [itex]\partial y/\partial v= -u/v^2= -u/(u^2/y^2)= -y^2/u[/itex].

    That is, [itex]df/dv= y\partial f/\partial u- (y^2/u)\partial f/\partial y[/itex]

    And, since here f(u,y)= u, [itex]\partial f/\partial u= 1[/itex] and [itex]\partial f/\partial y= 0[/itex] so we have df/dv= y.
     
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