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Homework Help: Explanation of this Partial?

  1. Dec 11, 2006 #1
    [tex]\frac{\partial_P}{\partial_y}(2ysinxcosx-y+2y^2e^{(xy^2)}[/tex]

    I worked the first part no problem, but the second part I needed a little help from my calculator. This is what I got:

    [tex]2sinxcosx-1+4ye^{(xy^2)}[/tex]

    My question is, why does the partial of [tex]2y^2e^{(xy^2)}[/tex] come out to [tex]4ye^{(xy^2)}[/tex]?

    I see the where the 4y comes from, but how come the [tex]e^{(xy^2)}[/tex] stays exactly the same.

    I also know that [tex]\frac{d}{dx}(e^x) = e^x[/tex]

    Thanks!
     
    Last edited: Dec 11, 2006
  2. jcsd
  3. Dec 11, 2006 #2

    dextercioby

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    Science Advisor
    Homework Helper

    It's a mistake in the last term of the sum.

    [tex] \frac{\partial}{\partial y} 2y^2 e^{xy^2} [/tex]

    should be done using product rule and chain rule.

    Daniel.
     
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