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Math Help - Partial diff. is this correct

  1. #1
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    Partial diff. is this correct

    Hi everyone,

    Need help to verify this working.Thank you for all help & support.

    Find \frac{dz}{dx} & \frac{dz}{dy} for
    x^2y^3z+yz^2-xy=0
    2xy^3z+x^2y^3\frac{dz}{dx}+2yz\frac{dz}{dx}-y=0
    (x^2y^3+2yz)\frac{dz}{dx}=y-2xy^3z
    \frac{dz}{dx}=\frac{y-2xy^3z}{x^2y^3+2yz}

    3x^2y^2z+x^2y^3\frac{dz}{dy}+z^2+2yz\frac{dz}{dy}-x=0
    \frac{dz}{dy}=\frac{x-3x^2y^2z-z^2}{x^2y^3+2yz}

    Thank you in advance for all help & support,really appreciate.
    Last edited by anderson; March 11th 2010 at 03:06 AM.
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  2. #2
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    What is the right hand side of this function? Is it set =0?
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  3. #3
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    Hi ProveIt,

    yes, it's equal to 0.
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  4. #4
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    x^2y^3z + yz^2 - xy = 0

    x^2y^3z + yz^2 = xy

    z^2 + x^2y^2z = x

    z^2 + x^2y^2z + \left(\frac{1}{2}x^2y^2\right)^2 = \left(\frac{1}{2}x^2y^2\right)^2 + x

    \left(z + \frac{1}{2}x^2y^2\right)^2 = \frac{1}{4}x^4y^4 + x

    z + \frac{1}{2}x^2y^2 = \pm \sqrt{\frac{1}{4}x^4y^4 + x}

    z + \frac{1}{2}x^2y^2 = \pm \sqrt{\frac{x^4y^4 + 4x}{4}}

    z + \frac{1}{2}x^2y^2 = \pm \frac{\sqrt{x^4y^4 + 4x}}{2}

    z = \frac{-x^2y^2 \pm \sqrt{x^4y^4 + 4x}}{2}.


    Now try to differentiate with respect to x and y.
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  5. #5
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    thank you for replying.there is a different alternative to answer this question though.
    Last edited by anderson; March 11th 2010 at 07:04 AM.
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