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Homework Help: GRE Partial Derivatives

  1. Aug 21, 2007 #1
    I am reading "Cracking the GRE Math Subject Test - Princeton Review, 3rd Ed." and and confused by the section on the chain rule for partial derivatives. The method in the book is as follows:

    1) Draw a diagram to show how the variables depend on each other, with an arrow meaning "depends on"

    2) To find a derivative, find all paths from the dependent variable to the independent variable. Each path gives a product of partial derivatives.

    3) Sum all products from different paths in 2.

    I thought I understood this (and I think I have paraphrased it correctly!) but they give the following example of a complicated situation:

    Let z = F(u, v, y), where u = f(v, x) and v = g(x, y). We are to find [tex]\frac{\partial z}{\partial y}[/tex]. They provide the following diagram, which I have reproduced (in glorious code-o-vision):

    Code (Text):

      |   ^ \     /
      |   || \   /
      |   ||  \ /
    --z   ||   \
    | |   ||  / \
    | |   || /   \
    | |   |v/     \
    | ---->v------->y
    |               ^

    Arrows from z to u, v and y
    Arrows from u to x, y and v
    Arrows from v to x, y and u

    They then give the answer

    [tex]\frac{\partial z}{\partial y} = \frac{\partial z}{\partial y} + \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial u}\frac{\partial u}{\partial y}[/tex]

    My problems:

    1) Does their diagram have two too many arrows? (I think the arrows from v to u and from u to y are wrong.) If their arrows are correct, why are they there?

    2) Using their method, there are paths from z to y they have not considered.

    3) I get the answer (using my diagram)

    [tex]\frac{\partial z}{\partial y} = \frac{\partial z}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial z}{\partial u}\frac{\partial u}{\partial v}\frac{\partial v}{\partial y}[/tex]

    are these the same? (I don't think so.)

    Thanks. I hope this is clear.
  2. jcsd
  3. Aug 21, 2007 #2
    Your answer seems correct to me. They probably made a typo. I mean if u = f(v,x) has no y terms in it, then it makes no sense to take the partial derviative of u with respect to y
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