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Partial derivation question

  1. Feb 8, 2012 #1
    From Mary Boas's book, chapter 4.

    Z = x^3 -e^xy


    1- Z(x) = 3x^2 y - ye^xy make sense according to derivation rule d(e^u)/dx = e^u.u`
    2-Z(y) = 3^x - xe^xy make sense too.

    3-Z(y)x)) = 3x^2 - e^xy - xye^xy where e^xy came from ??




    Thank you.
     
  2. jcsd
  3. Feb 9, 2012 #2
    Even though this is one of the first examples in chapter 4 a more complete reference would have helped get you a quicker response.

    [tex]\begin{array}{l}
    \frac{{\partial f}}{{\partial x}} = 3{x^2}y - y{e^{xy}} \\
    \frac{{\partial f}}{{\partial x\partial y}} = 3{x^2} - {e^{xy}} - xy{e^{xy}} \\
    \end{array}[/tex]

    So you understand how to take the first partial differential with respect to x or y.

    I have shown the one with respect to x.

    The second partial that you are having trouble with is formed by taking either the first differential with respect to x and differentiating it with respect to y or the other way round. The result is the same.

    I have chosen to take the first partial with respect to x and then differentiate it with respect to y.

    The first partial is formed from two terms

    1)
    [tex]3{x^2}y[/tex]

    from which we get 3x2 since x is considered constant in this second differentiation.

    2)[tex]y{e^{xy}}[/tex]

    Which is the product of two functions of y, viz y and exy

    This will yield two terms according to the product rule,

    differentiating y with respect to -y yields 1 and so multiplied by exy yields the second term -exy

    differentiating exy with respect to y yields xexy and so multiplied by -y yields -xyexy

    which is the third term you are having trouble with. does this help?
     
  4. Feb 10, 2012 #3
    It make perfec sense.



    Thank you very much!!
     
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