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Partial Derivatives Instead of Implicit

  1. May 13, 2005 #1
    On MathWorld's site, they said that

    [tex](\frac{\partial{y}}{\partial{x}}){_f} = -\frac{(\frac{\partial{f}}{\partial{x}})_{y}}{(\frac{\partial{f}}{\partial{y}})_{x}}[/tex]

    So can this method be used instead of implicit differentiation? Will I get the same result? This seems kind of like a parametrics process if this is true.

    Thanks,
    Jameson
     
    Last edited: May 13, 2005
  2. jcsd
  3. May 13, 2005 #2

    dextercioby

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    Yep,that's the formula from the statement of the theorem of implicit functions.

    Daniel.
     
  4. May 13, 2005 #3
    Is that for f(x,y)? If so isnt the right hand side = -1?

    y(f,x)?
     
  5. May 13, 2005 #4
    yes, and there is nothing mysterious about this equation, btw:

    for [itex]f(x,y) = 0[/itex]

    [tex] df = \frac{\partial{f}}{\partial{x}}dx + \frac{\partial{f}}{\partial{y}}dy = 0[/tex]

    [tex]\frac{\partial{f}}{\partial{x}}dx = -\frac{\partial{f}}{\partial{y}}dy[/tex]

    [tex]\frac{dy}{dx} = -\frac{\frac{\partial{f}}{\partial{x}}}{\frac{\partial{f}}{\partial{y}}}[/tex]
     
  6. May 13, 2005 #5
    Quetzalcoatl9 I'm not sure if you addressed my question. Isnt there a theorem that says the double partial of a function with respect to alternating variables are equal?

    [tex] f(x,y), f_{xy} = f_{yx} [/tex] Right? Isn't that whats going on here?
     
  7. May 13, 2005 #6
    I don't understand what you are asking, whozum...do you mean to ask if partial derivatives commute?
     
  8. May 13, 2005 #7
    I dont know what you mean by commute, but please verify that

    [tex]For \ f(x,y), \ f_{xy} = f_{yx} [/tex]

    If so, then please explain why the right hand side of the original OP's equation isnt just -1
     
  9. May 14, 2005 #8

    Gza

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    where'd you get the idea the right side was -1?
     
  10. May 14, 2005 #9
    [tex] \left(\frac{\delta f}{\delta x}\right)_y = f_{xy} [/tex]

    [tex] \left(\frac{\delta f}{\delta y}\right)_x = f_{yx} [/tex]

    [tex] f_{yx} = f_{xy} [/tex]

    edit: I'm obviously missing something, I just want someone to point out what it is.
     
    Last edited: May 14, 2005
  11. May 14, 2005 #10

    dextercioby

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    Yes,you are.Missing something,that is.We use this notation

    [tex] \left (\frac{\partial f}{\partial x}\right)_{y}\equiv \frac{\partial f(x,y)}{\partial x} [/tex]

    ,that is we explain which variables are kept constant during the partial differentiaition,viz. "y" in this case.

    This notation,or convention,if u prefer,is very common in thermodynamics.

    Daniel.
     
  12. May 14, 2005 #11
    So that hanging y on the LHS is just to indicate that its constant, then. I had that confused as a derivative.
     
  13. May 14, 2005 #12

    arildno

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    The notation is excellent in a field like thermo-dynamics, where it is convenient to switch between which quantities are to be regarded as independent quantities and which are to be regarded as dependent quantities.

    It is rather redundant in a field where there exisst a "natural" choice of the independent variables.
     
  14. May 14, 2005 #13

    dextercioby

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    To give u an idea

    [tex] \frac{\partial F}{\partial V} [/tex]

    ,where F is the free energy/Helmholtz potential & V is the volume,doesn't mean anything in thermodynamics.

    Daniel.
     
  15. May 14, 2005 #14

    arildno

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    Are you talking to me??
     
  16. May 14, 2005 #15

    dextercioby

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    Of course,not.I gave that example to Whozum...

    Hmmmmm...:tongue2:

    Daniel.
     
  17. May 14, 2005 #16
    ...<in bronx-accented Al Pacino voice> :rofl:
     
  18. May 14, 2005 #17

    arildno

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    I haven't bothered to get a driving license yet.
    Perhaps I should..:devil:
     
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