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Math Help - multivaribles 3??? caculus

  1. #1
    Member zangestu888's Avatar
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    multivaribles 3??? caculus

    let f(x,y,z)=2x^2y^2z + 4xy^3z^2 find fxyz(2,1,1)

    do i have to find partial dervtive of fx and fy and fz and sub the value in?
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  2. #2
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    Quote Originally Posted by zangestu888 View Post
    let f(x,y,z)=2x^2y^2z + 4xy^3z^2 find fxyz(2,1,1)

    do i have to find partial dervtive of fx and fy and fz and sub the value in?
    Is this what you're trying to find?

    f_{xyz}(2, 1, 1)?
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  3. #3
    MHF Contributor arbolis's Avatar
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    I don't think so. You would have to find f_x then partial derivate it with respect to y. Once you get this function, partial derivate with respect to z.
    Then evaluate this function in (2,1,1).
    This method however doesn't work for all functions. For example the function f(x,y)=xy \frac{x^2-y^2}{x^2+y^2}.
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  4. #4
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    In other words, you're finding

    \frac{\partial}{\partial z}\left(\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x}\left(f(x, y, z)\right)\right)\right)

    and then evaluating it at x = 2, y = 1, z = 1.
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  5. #5
    Member zangestu888's Avatar
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    Yeah the book says find
    <br />
f_{xyz}(2, 1, 1)<br />

    does that mean i find fx then fxy then fxyz then find it at the point
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  6. #6
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    Quote Originally Posted by zangestu888 View Post
    Yeah the book says find
    <br />
f_{xyz}(2, 1, 1)<br />

    does that mean i find fx then fxy then fxyz then find it at the point
    It certainly does.
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