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Math Help - partial differentiation problem

  1. #1
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    partial differentiation problem

    Hi

    can anyone help me partially differentiate

    u = (-2xyz)/(x^2 + y^2)^2

    Partial du/dx = ?

    and

    v = (z(x^2-y^2))/(x^2+y^2)^2

    Partial dv/dx = ?

    Any ideas for the second partial derivatives?

    Any help would be greatly appreciated

    Many thanks

    zentus
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, zentus!

    Partial derivatives

    u \:= \:\frac{-2xyz}{(x^2 + y^2)^2} . . . Find \frac{\partial u}{\partial x}

    To differentiate with respect to x, treat the other variables as constants.


    \frac{\partial u}{\partial x} \;=\;\frac{(x^2+y^2)\cdot(-2yz) - (-2xyz)\cdot2(x^2+y^2)\cdot2x}{(x^2+y^2)^4} \;= . \frac{-2yz(x^2+y^2)^2 + 8x^2yz(x^2+y^2)}{(x^2+y^2)^4}

    Factor: . \frac{-2yz(x^2+y^2)\cdot\left[(x^2+y^2) - 4x\right]}{(x^2+y^2)^4}

    Reduce: . \frac{\partial u}{\partial x} \;=\;\frac{-2yz(x^2+y^2 - 4x)}{(x^2+y^2)^3}


    Now try the second one yourself . . .

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  3. #3
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    thank you!

    Soroban

    Thank you!
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