Results 1 to 6 of 6

Thread: multivaribles 3??? caculus

  1. #1
    Member zangestu888's Avatar
    Joined
    Jan 2009
    Posts
    91

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    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?

    $\displaystyle f_{xyz}(2, 1, 1)$?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    I don't think so. You would have to find $\displaystyle f_x$ then partial derivate it with respect to $\displaystyle y$. Once you get this function, partial derivate with respect to $\displaystyle z$.
    Then evaluate this function in $\displaystyle (2,1,1)$.
    This method however doesn't work for all functions. For example the function $\displaystyle f(x,y)=xy \frac{x^2-y^2}{x^2+y^2}$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    In other words, you're finding

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

    and then evaluating it at $\displaystyle x = 2, y = 1, z = 1$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member zangestu888's Avatar
    Joined
    Jan 2009
    Posts
    91
    Yeah the book says find
    $\displaystyle
    f_{xyz}(2, 1, 1)
    $

    does that mean i find fx then fxy then fxyz then find it at the point
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by zangestu888 View Post
    Yeah the book says find
    $\displaystyle
    f_{xyz}(2, 1, 1)
    $

    does that mean i find fx then fxy then fxyz then find it at the point
    It certainly does.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Multivariable Caculus Leverl Curves
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Apr 21st 2009, 06:38 PM
  2. Caculus Genius wanted,,
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 8th 2009, 08:15 AM
  3. Differntial Caculus Salt Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 24th 2009, 05:08 PM
  4. polynomial caculus
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 28th 2008, 02:13 PM
  5. Caculus Homework Help!!
    Posted in the Calculus Forum
    Replies: 14
    Last Post: Aug 6th 2006, 02:32 PM

Search Tags


/mathhelpforum @mathhelpforum