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Multivariable chain rule (with implicit variable)

  1. May 3, 2010 #1
    1. The problem statement, all variables and given/known data
    If T is implicitly defined via the relationship f(x, y, z, T) = 0 to be a differentiable function of x, y and z, show that the first partial derivative of T with respect to z can be found using:

    [tex]
    \frac{\partial T}{\partial z} = -\frac{\partial f}{\partial z} / \frac{\partial f}{\partial T}
    [/tex]

    2. Relevant equations
    This is a similar example from my textbook, for f(x, y, z) = 0, where z is assumed to be a function of x only.

    [tex]
    \frac{\partial f}{\partial x} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x}
    [/tex]

    [tex]
    0 = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x}[/tex] as x and y are independent variables, meaning the derivative of y with respect to x is 0.

    [tex]
    \frac{\partial z}{\partial x} = -\frac{\partial f}{\partial x} / \frac{\partial f}{\partial z}
    [/tex]


    3. The attempt at a solution
    I tried using something similar for the question and it happened to work out, but I'd like to actually understand it properly. It's mostly the implicit part that I'm unsure of for this question. Here's what I did for it:

    [tex]
    \frac{\partial f}{\partial x} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial z} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial z} + \frac{\partial f}{\partial T}\frac{\partial T}{\partial z}
    [/tex]

    [tex]
    0 = \frac{\partial f}{\partial z} + \frac{\partial f}{\partial T}\frac{\partial T}{\partial z}[/tex] as x and y are not dependent on z, derivatives with respect to z must be 0 (not too sure about this part either)

    [tex]
    \frac{\partial T}{\partial z} = -\frac{\partial f}{\partial z} / \frac{\partial f}{\partial T}
    [/tex]

    It gets the right answer but I'm still a bit confused by it. I was also having trouble figuring out what the dependency tree for this relationship would look like.
    For the textbook example, it mentions that z is assumed to be a function of x only; I'm not sure if that applies to applies to this question in a similar way or if it's for both x and y (or maybe something completely different).
     
    Last edited: May 3, 2010
  2. jcsd
  3. May 3, 2010 #2

    tiny-tim

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    Homework Helper

    Welcome to PF!

    Hi Refraction! Welcome to PF! :smile:

    (have a curly d: ∂ :wink:)

    Your proof is fine.

    f is a function of 4 variables.

    3 of those are independent, and the 4th is a function of those three variables.

    If you change z slightly, that won't affect x or y, but it will affect z (obviously :rolleyes:) and T, so you must include ∂f/∂z and ∂f/∂T ∂T/∂z.
    No, it doesn't apply in this case … the only constraint is that T depends on x y and z.
     
  4. May 3, 2010 #3
    Awesome, thanks! This one question had been bugging me for the last week, your explanation helps a lot.

    The only other thing I was unsure of was the dependency tree (which isn't really necessary for this question anyway). I understand how they work for something like a function z = f(x, y), where x and y are both differentiable with respect to multiple variables (e.g. s and t):

    rwhveu.png

    since it's easy enough to figure out that (as an example) ∂z/∂s = ∂z/∂x ∂x/∂s + ∂z/∂y ∂y/∂s from this. But for something like this question, where it doesn't explicitly tell you that sort of thing, I'm not so sure what it would look like.
     
    Last edited: May 3, 2010
  5. May 3, 2010 #4

    tiny-tim

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    Homework Helper

    Hi Refraction! :smile:

    I've never come across dependency trees before. :redface:

    From your diagram, it seems that the dependency tree in this case would need T on a higher level than x y and z, so that lines could go down from f to x y z and T, and from T to x y and z.

    I honestly don't see the advantage of drawing that. :confused:
     
  6. May 3, 2010 #5
    Thanks again, that does sound like it would make sense. And thinking about it now, it doesn't look like it would have much of a use for this kind of question anyway.
     
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