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Thread: 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 $\displaystyle f(x, y, z)$, but I know that there exist a known relation between $\displaystyle x$, $\displaystyle y$ and $\displaystyle z$ that can be expressed as $\displaystyle z = z(x, y)$ (or, equivalently, a known function $\displaystyle g$ satisfying that $\displaystyle g(x, y, z) = 0$ with $\displaystyle \partial g/\partial z = 1$). Is there a way to write the derivative $\displaystyle \partial f/\partial z$ in terms of $\displaystyle \partial f/\partial x$, $\displaystyle \partial f/\partial y$, $\displaystyle z(x, y)$, $\displaystyle \partial z/\partial x$ and $\displaystyle \partial z/\partial y$? 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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