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Problem of Partials

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Find [tex]\frac{\partial{w}}{\partial{x}}[/tex] for [tex]F(x,y,z,w)=xyz+xzw-yzw+w^2-5=0[/tex]

    3. The attempt at a solution
    I know how to get partials (such as Dz/Dx) of functions in the form z = f(x,y). I'm having trouble figuring out this one since it's asking for the partial of w, which is all over the function. I tried putting everything equal to w, but that w2 doesn't help. I thought of perhaps doing implicit differentiation, but I still end up with that extra w on one side.

    I'm sure it's something tiny that I'm not seeing, but I need a little push in the right direction.

    Thank you much.
     
  2. jcsd
  3. Dec 4, 2009 #2
    Re: Partials

    Let me rewrite it for you:

    [tex] w^2 + w(xz-yx)+ (yz-5) = 0[/tex]

    Now if I also write this:
    [tex]w^2 + a*w + b = 0[/tex]

    Does it ring a bell what the next step should be?
     
  4. Dec 4, 2009 #3
    Re: Partials

    I tried that, but that would seem to only get everything factored, or to find the values of w when F(x,y,z) is 0. I could be wrong though. I tried something else, I figured since y and z are constants I rearranged it as:

    [tex](yz)x + (z)xw - (yz)w + w^2 - 5 = 0[/tex]

    Then I just took the partials of all that, and got:

    [tex]\frac{\partial{w}}{\partial{x}} = 1(yz) + (x+w)(z) - 1(yz) + 2w = 2w + zw + zx[/tex]

    I have no way to check if this is right, but hopefully it is.

    Thanks again.
     
  5. Dec 5, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Partials

    Use "implicit differentiation". For example, if [itex]F(x,y,w)= xw+ yw+ w^2= 0[/itex], assuming that w is a function of x and y but x and y are independent variables, then
    [tex]\frac{\partial F}{\partial x}= w+ x\frac{\partial w}{\partial x}+ y\frac{\partial w}{\partial x}+ 2w\frac{\partial w}{\partial x}= 0 [/tex]
    so
    [tex](x+ y+ 2w)\frac{\partial w}{\partial x}= -w[/tex]
    and
    [tex]\frac{\partial w}{\partial x}= \frac{-w}{x+ y + 2w}[/tex]
     
  6. Dec 5, 2009 #5
    Re: Partials

    I actually had tried implicit differentiation...I don't know why I thought it was wrong because of the w2. Anyways, I tried it again:

    Problem: [tex]F(x,y,z,w)=xyz+xzw-yzw+w^2-5=0[/tex]

    [tex]\frac{\partial{w}}{\partial{x}}=0+w+x\frac{\partial{w}}{\partial{x}}-yz\frac{\partial{w}}{\partial{x}}+2w\frac{\partial{w}}{\partial{x}}= \frac{-w}{x-yz+2w}[/tex]

    Thank you all for your help.
     
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