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Homework Help: Partial derivative concept

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Given the partial derivative df/dx= 3-3(x^2)

    what is d^2f/dydx?

    I'm not sure if the answer would be 0, since x is held constant, or if it would remain 3-3(x^2) (since df/dx is a function of x now?)
     
  2. jcsd
  3. Apr 27, 2008 #2

    G01

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    The answer is one of those choices. Here, think about it like this:

    You are given a function:

    [tex]g(x)=3-3x^2[/tex]

    You want to find: [tex]\frac{\partial g}{\partial y}[/tex]

    What is that derivative? Now, what if: [tex]g(x)=\frac{\partial f}{\partial x}[/tex]

    Does this change the partial derivative of g with respect to y?
     
  4. Apr 27, 2008 #3

    Dick

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    You were right the first time. With x held constant the d/dy is just differentiating a constant. It's 0.
     
  5. Apr 27, 2008 #4

    HallsofIvy

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    As is always true, with "nice" functions, the two mixed derivatives are equal. You could find [itex]\partial^2 f/\partial x\partial y[/itex] by differentiating first with respect to x, then with respect to y: first getting -6x and then, since it does not depend on y, 0. Or you could differentiate first with respect to y, then with respect to x: getting 0 immediately and then, of course, the derivative of "0" with respect o x is 0.

    I, and I suspect many who read your post, was momentarily taken aback since I thought you were "holding x constant" through both derivatives. But you are correct: since this function does not depend on y, any derivative of it with respect to y, is 0.
     
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