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Math Help - Generalized chain rule

  1. #1
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    Generalized chain rule

    Hi! This problem is somewhat related to a thread that I have posted previously: Theorem considering multivariable calculus
    The problem that I have now is the following: Say that I have an unknown function f(x, y, z), but I know that there exist a known relation between x, y and z that can be expressed as z = z(x, y) (or, equivalently, a known function g satisfying that g(x, y, z) = 0 with \partial g/\partial z = 1). Is there a way to write the derivative \partial f/\partial z in terms of \partial f/\partial x, \partial f/\partial y, z(x, y), \partial z/\partial x and \partial z/\partial y? If not, why not?

    Thanks for any help,
    Simon
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  2. #2
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    Re: Generalized chain rule

    It has been a long times since I have done multivariable calculus, so I may have this wrong, but I believe:

    $\dfrac{df}{dz} = 0 = \dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial z} + \dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial z} + \dfrac{\partial f}{\partial z}$

    Solving for $\dfrac{\partial f}{\partial z}$ gives

    $\dfrac{\partial f}{\partial z} = -\dfrac{ \left( \dfrac{\partial f}{\partial x} \right) }{ \left( \dfrac{\partial z}{\partial x} \right) } - \dfrac{ \left( \dfrac{\partial f}{\partial y} \right) }{ \left( \dfrac{\partial z}{\partial y} \right) }$
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  3. #3
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    Re: Generalized chain rule

    Quote Originally Posted by SlipEternal View Post
    It has been a long times since I have done multivariable calculus, so I may have this wrong, but I believe:

    $\dfrac{df}{dz} = 0 = \dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial z} + \dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial z} + \dfrac{\partial f}{\partial z}$

    Solving for $\dfrac{\partial f}{\partial z}$ gives

    $\dfrac{\partial f}{\partial z} = -\dfrac{ \left( \dfrac{\partial f}{\partial x} \right) }{ \left( \dfrac{\partial z}{\partial x} \right) } - \dfrac{ \left( \dfrac{\partial f}{\partial y} \right) }{ \left( \dfrac{\partial z}{\partial y} \right) }$
    pretty sure you have to put in the condition that the Jacobian is non-singular to be able to do this.
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  4. #4
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    Re: Generalized chain rule

    Quote Originally Posted by romsek View Post
    pretty sure you have to put in the condition that the Jacobian is non-singular to be able to do this.
    Ah, it has been a long time... You are correct. My bad.
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  5. #5
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    Re: Generalized chain rule

    By the time I got it, it was done. But anyhow, with apologies, and a little more detail:

    df=fxdx+fydy+fzdz=0
    fz=-fxxz-fyyz

    z=z(x,y)
    dz=zxdx+zydy
    zxdx=dz-zydy
    zxxz=1
    xz=1/zx
    similarly
    yz=1/zy

    fz=-fx/zx-fy/zy
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