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Math Help - Partial Derivatives

  1. #1
    Junior Member
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    Jan 2008
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    Partial Derivatives

    Got another question, this time its a bit harder and im not sure how to get the partial derivatives
    The function is:
    f(x,y) = e^1/2xy

    e to the power of one half(x times y)

    Need help with finding:
    fx
    fy
    fxx
    fyy
    fxy
    fyx

    Thanks!
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Hockey_Guy14!

    Find the first and second partial derivatives:

    . . . f(x,y) \:=\: e^{\frac{1}{2}xy}

    . . \begin{array}{ccccc}f_x &=& e^{\frac{1}{2}xy}\cdot\frac{1}{2}y &=& \frac{1}{2}y\,e^{\frac{1}{2}xy} \\ \\[-3mm]<br />
f_y &=& e^{\frac{1}{2}xy}\cdot\frac{1}{2}x &=& \frac{1}{2}x\,e^{\frac{1}{2}xy} \end{array}


    . . \begin{array}{ccccc}f_{xx} &=& \frac{1}{2}y\,e^{\frac{1}{2}xy}\cdot\frac{1}{2}y &=& \frac{1}{4}y^2\,e^{\frac{1}{2}xy} \\ \\[-3mm]<br />
f_{yy} &=& \frac{1}{2}x\,e^{\frac{1}{2}xy}\cdot\frac{1}{2}x &=& \frac{1}{4}x^2\,e^{\frac{1}{2}xy} \end{array}


    . . \begin{array}{ccccc}<br />
f_{xy} &=& \frac{1}{2}y\cdot e^{\frac{1}{2}xy}\cdot\frac{1}{2}x + \frac{1}{2}\,e^{\frac{1}{2}xy} & = &\frac{1}{4}e^{\frac{1}{2}xy}\,(xy + 2) \\ \\[-3mm]<br />
f_{yx} &=& \frac{1}{2}x\,e^{\frac{1}{2}xy}\cdot\frac{1}{2}y + \frac{1}{2}e^{\frac{1}{2}xy} &=& \frac{1}{4}e^{\frac{1}{2}xy}\,(xy + 2)<br />
\end{array}

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