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How Partial Derivative Changing Variable Formula works ?

  1. Jul 26, 2012 #1
    1. The problem statement, all variables and given/known data
    The changing variable formula in partial derivative
    f(u,v)
    x=x(u,v)
    y=y(u,v)
    (∂f/∂x)y = (∂f/∂u)v(∂u/∂x)y + (∂f/∂v)u(∂v/∂x)y
    I khow the how chain rule works, but I don't know why in the (∂f/∂u) v is constant and in the (∂u/∂x) y is constant
    2. Relevant equations
    3. The attempt at a solution
     
  2. jcsd
  3. Jul 26, 2012 #2

    tiny-tim

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    welcome to pf!

    hi pyfgcr! welcome to pf! :smile:
    that's what a partial derivative is

    it's defined as being calculated with all the other variables kept constant

    so it actually depends on what the other variables are: for example, if a function f is expressed both as f(x,y,z) and f(r,z,θ) (ie cartesian coordinates and cylindrical coordinates),

    then, even though the z is the same, ∂f/∂z is different in each case

    (in practice you'll probably avoid confusion by using different letters for the function, f(x,y,z) and g(r,z,θ) … but if you don't, you will need to write either (∂f/∂z)x,y or (∂f/∂z)r,θ :wink:)
     
  4. Jul 26, 2012 #3
    But x is express in u and v, whereas in ∂u/∂x, y is kept constant.
    What explain it ?
     
  5. Jul 26, 2012 #4

    HallsofIvy

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    Again, that is the definition of the partial derivative.

    The partial derivative, with respect to x, of f(x,y) at [itex](x_0, y_0)[/itex] is defined as
    [tex]\lim_{h\to 0}\frac{f(x_0+h, y_0)- f(x_0,y_0)}{h}[/tex]
    and the partial derivative with respect to y is
    [tex]\lim_{y\to 0}\frac{f(x_0, y_0+h)- f(x_0,y_0)}{h}[/tex]

    When taking the derivative with respect to one variable, all other variables are held constant.
     
  6. Jul 26, 2012 #5
    So in (∂x/∂u)y, by definition:
    (∂x)/(∂u)y=[itex]lim_{h\rightarrow0}[/itex][itex]\frac{x(u_{0},y_{0}+h)-x(u_{0},y_{0})}{h}[/itex]
    A little awkward, since x=x(u,v), I think

    In equation: (∂f/∂x)y = (∂f/∂u)v(∂u/∂x)y + (∂f/∂v)u(∂v/∂x)y
    Why in the term (∂f/∂u)v, v is kept constant, not y.
     
  7. Jul 26, 2012 #6
    On a realistic level, with the variables you've been given, you're going to need to invert something to get [itex]u = u(x,y)[/itex] and the same for [itex]v[/itex].
     
  8. Jul 27, 2012 #7
    Re: welcome to pf!

    And in (∂u/∂x)y, with x, "the other variable" than u is v, but y is kept constant, not v
     
  9. Jul 27, 2012 #8

    tiny-tim

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    hi pyfgcr! :smile:
    no

    x and y are functions of u and v

    u and v are functions of x and y

    in particular, u is a function of x and y

    so, for u, the "other variable" than x is y

    so ∂u/∂x keeps y constant :wink:
     
  10. Jul 27, 2012 #9
    Thanks a lot!
     
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