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Partial Derivatives

  1. Jul 1, 2007 #1
    Find the four second partial derivatives for f(x,y) = y^2e^x + xycosx

    I am stuck on the last part... here's what I got so far:

    Zx = y^2e^x - ysinx
    Zxx = y^2e^x - ycosx
    Zxy = ?

    Zy = 2ye^x + xcosx
    Zyy = 2e^x + xcosx
    Zyx = ?

    I need help with solving for xy. Both should end up being the same. :eek:
  2. jcsd
  3. Jul 1, 2007 #2


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    Homework Helper

    the trick is to ignore the other variable whenever you are not differentiating with respect to it. eg.
    [tex]\frac{\partial}{\partial x} x e^y = e^y[/tex] and
    [tex]\frac{\partial}{\partial y} x e^y = x e^y[/tex]

    in first case you treat y as if it is a constant whereas in the second you treat x as constant instead. the same rules apply when you go to higher derviatives
    [tex]\frac{\partial}{\partial x} \frac{\partial}{\partial y} (x e^y)
    = \frac{\partial}{\partial x}(x e^y) = e^y[/tex]
  4. Jul 1, 2007 #3


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    Homework Helper

    by the way your Zx is wrong
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