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Homework Help: Evalute partial diff with values of x&y

  1. Feb 20, 2012 #1
    Have worked through to get some sort of answer, but find the evaluation (simplest bit) impossible - wood for trees, I think.

    given d2(e^(xy^2)/dy^2 I need to evaluate for x=2, y=0

    now using chain rule d(e^(xy^2))/dy = d(e^u)/du du/dy
    where u=xy^2 & d(e^u)/du = e^u
    => x(2y) e^(xy^2)
    fine for 1st order

    moving on
    d(x(2y) e^(xy^2))/dy

    removing constants

    use product rule
    d(uv)/dy = v du/dy + u dv/dy
    u =e^(xy^2) & v=y

    => 2x(e^(xy^2)(d(y)/dy)+y(d(e^xy^2)))

    chain rule again
    where u=xy^2 & d(e^u)/du = e^u


    which shuld be d2(e^xy^2)/dy2
    => 2x(xy(2y)e^(xy^2)+e^(xy^2)

    Now evaluate for x=2, y=0

    everything goes to 0 ??????

    I'v gone throught he difficult bits in the book, now the simple stumps me!!!
    Please help before I bang my head against the wall again!!!
  2. jcsd
  3. Feb 20, 2012 #2


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    Homework Helper

    \frac{\partial}{\partial y}e^{xy^2} = 2xye^{xy^2}

    looks good so far

    \frac{\partial^2}{\partial^2 y}e^{xy^2}
    = \frac{\partial}{\partial y}2xye^{xy^2}
    = \left( \frac{\partial}{\partial y}2xy \right) e^{xy^2} +2xy\frac{\partial}{\partial y}e^{xy^2}
    then you know what the value of the second derivative in the RHS is, as you have already calculated it

    below I think you have lost some brackets, that you may want to check

    if you put the extra backet in and your final expression is
    2x(xy(2y)e^(xy^2)) + e^(xy^2)

    then knowing e^0=1, this expression is non-zero
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