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Chain rule substitution help

  1. Sep 7, 2005 #1
    Let f: [tex] \Re^3 \rightarrow \Re [/tex] be differentiable. Making the substitution

    [tex] x = \rho \cos{\theta} \sin{\phi}, y = \rho \sin{\theta} \sin{\phi}, z = \rho \cos{\phi} [/tex]

    (spherical coordinates) into f(x,y,z), compute (partially) df/d(rho), df/d(theta), and df/d(phi) in terms of df/dx, df/dy, and df/dz.

    I'm just not sure I understand the question. Does it involve pulling out a very long chain rule?
    Last edited: Sep 7, 2005
  2. jcsd
  3. Sep 7, 2005 #2


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    It involves the chain rule, not sure what you mean about the very long part.
    [tex]\frac{\partial f}{\partial\rho}=\frac{\partial f}{\partial x} \ \frac{\partial x}{\partial\rho}+\frac{\partial f}{\partial y} \ \frac{\partial y}{\partial\rho}+\frac{\partial f}{\partial z} \ \frac{\partial z}{\partial\rho}[/tex]
    [tex]\frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x} \ \frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y} \ \frac{\partial y}{\partial\theta}+\frac{\partial f}{\partial z} \ \frac{\partial z}{\partial\theta}[/tex]
    [tex]\frac{\partial f}{\partial\phi}=\frac{\partial f}{\partial x} \ \frac{\partial x}{\partial\phi}+\frac{\partial f}{\partial y} \ \frac{\partial y}{\partial\phi}+\frac{\partial f}{\partial z} \ \frac{\partial z}{\partial\phi}[/tex]
    The general form of the chain rule being
    [tex]\frac{\partial f}{\partial x}=\sum_{k=1}^n \frac{\partial f}{\partial u_k} \ \frac{\partial u_k}{\partial x}[/tex]
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