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Partial Differentiation

  1. Mar 24, 2008 #1
    f(x,y)=2e^(x^2y), then fx(0,1) = ?

    fx = 4xye^(x^2y)

    I'm told that I'm getting this question wrong but don't know how. Can anybody please help show me what I'm doing wrong on this one?
  2. jcsd
  3. Mar 24, 2008 #2

    Gib Z

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    You are doing partial differentiation with respect to x, so you can treat y as a constant.

    It's just simply using the chain rule and [itex](e^x)' = e^x[/itex].

    If you do that correctly you should get up to [itex] 2 \cdot ( 2y \cdot x^{2y-1} ) \exp (x^{2y})[/itex].

    The x term is a nice fat zero, giving the final result of 0.
  4. Mar 24, 2008 #3
    Crap! I see now. Thanks Gib.
  5. Mar 24, 2008 #4


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    Is that
    as Gib Z assumed or
    which is what I would assume?
    If it is the latter then
    [tex]f_x= 4xye^{x^2y}[/tex]
    as you have. Then [itex]f_x(0,1)= 4(0)(1)e^{(0)(1)}= 0[/itex], not 1, because of the "0" multiplying the exponential.
  6. Mar 24, 2008 #5

    Gib Z

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    Damn i just realised that :( Looking at his working it looks like you're right Halls =] Sorry guys.
  7. Mar 24, 2008 #6
    It was the latter. But I realized what I was doing wrong from Gib's reply. Thanks guys!
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