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Math Help - Partial Derivative

  1. #1
    Senior Member Danneedshelp's Avatar
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    Partial Derivative

    Hey, I am really rusty on my multivariable calc stuff and cannot seem to get the right answer to...

    f(x,y,z)=\tan^{-1}(\frac{1}{xy^{2}z^{3}}) with respect to x and with respect to y.

    Help? I don't know what techniques to imply. I just went ahead and did it as if it were just a single variable function since I only needed it with respect to x and also with respect to y.

    Thanks
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  2. #2
    Super Member malaygoel's Avatar
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    If you will show your work here..then it will be easier to tell
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  3. #3
    Super Member Random Variable's Avatar
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     \frac{\partial f}{\partial x} = \frac {1}{1+(\frac{1}{xy^{2}z^{3}})^{2}}* \frac {\partial}{\partial x} \frac{1}{ xy^{2}z^{3}}

     = \frac {1}{1+\frac{1}{x^{2}y^{4}z^{6}}}* \frac{\text{-}1}{ x^{2}y^{2}z^{3}}

     = \frac {-1}{x^{2}y^{2}z^{3}+ \frac{1}{y^{2}z^{3}}}

     = \frac {-y^{2}z^{3}}{x^{2}y^{4}z^{6} + 1}
    Last edited by Random Variable; June 30th 2009 at 07:20 PM.
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  4. #4
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    f(x,y,z)=\tan^{-1}(\frac{1}{xy^{2}z^{3}})

    You want \frac{\partial f}{\partial x} and \frac{\partial f}{\partial y}

    u = \frac{1}{xy^{2}z^{3}}

    \frac{\partial}{\partial u}(\frac{1}{tan{(u)}})

    Then when when calculating u' remember that

    \frac{\partial}{\partial x}(xy^{2}z^{3})=1*y^{2}z^{3}

    and that

    \frac{\partial}{\partial y}(xy^{2}z^{3})=2xyz^{3}

    As long as this isn't an implicit function, you treat the other variables as constants while deriving for one.

    Also remember the derivatives rules for fractions

    (\frac{f}{g})'=\frac{f'*g-f*g'}{g^{2}}

    And the derivative of tangent

    tan(u)' = u'*sec^{2}(u)

    That's all you need to know.
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