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Math Help - partial derivatives

  1. #1
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    partial derivative

    show that fxy =fyx for f(x,y)= x sqrt(y-x^2)

    when i find do this they are not equal but from the definition they are supposed to be...can some one show me the work to compare it to mine so i can see where i messed up???
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chris25 View Post
    show that fxy =fyx for f(x,y)= x sqrt(y-x^2)

    when i find do this they are not equal but from the definition they are supposed to be...can some one show me the work to compare it to mine so i can see where i messed up???
    f(x, y) = x \sqrt{y - x^2}

    f_x = \sqrt{y - x^2} + x \cdot \frac{1}{2 \sqrt{y - x^2}} \cdot -2x

    f_x = \sqrt{y - x^2} - \frac{x^2}{\sqrt{y - x^2}}

    f_{xy} = \frac{1}{2\sqrt{y - x^2}} + \frac{x^2}{2(y - x^2)^{3/2}}

    You do f_{yx}.

    -Dan
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  3. #3
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    partial derivatives

    for f(x,y) = x sqrt (y-x^2)
    find fy and fyx

    i think im messing up fy which is messing up fyx can anyone help me differentiate these???
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by chris25 View Post
    for f(x,y) = x sqrt (y-x^2)
    find fy and fyx

    i think im messing up fy which is messing up fyx can anyone help me differentiate these???
    how did you do it?
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  5. #5
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    used chain rule and received x/2sqrty-x^2
    after that i go wrong
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chris25 View Post
    used chain rule and received x/2sqrty-x^2
    after that i go wrong
    That is correct for f_y. So now do your x derivative. Use the quotient rule:
    f_{yx} = \frac{(x)' \cdot (2\sqrt{y - x^2)} - (x) \cdot (2 \sqrt{y - x^2})'}{(2 \sqrt{y - x^2})^2}
    where the primes represent derivatives with respect to x.

    -Dan
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chris25 View Post
    show that fxy =fyx for f(x,y)= x sqrt(y-x^2)

    when i find do this they are not equal but from the definition they are supposed to be...
    Is it always supposed to be? Consider D where D\equiv\left\{(x,y):y<x^2\right\}
    Last edited by Mathstud28; July 8th 2008 at 11:59 AM.
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