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Math Help - quick question on multivariable calc chain rule problem

  1. #1
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    quick question on multivariable calc chain rule problem

    find g of f at the indicated point a.
    g(x,y)= (x^2 * 3y, 3x-y^2)
    f(x1, x2, x3)= (x1x3, (x2)/(x1x3) )
    a= (3,1,-1)

    so the matrix for D(g) is
    2xy^3 3x^2 y^2
    3 -2y

    D(g(f(x1,x2,x3) is
    (2(x2)^3)/(x1x3)^2 3(x2)^2
    3 (-2x2)/(x1x3)

    what is D(f) ?
    i started with
    x3 0 x1
    but i dont know about the next row or if i even did the first row right.

    any help? thanks so much
    Last edited by holly123; September 28th 2009 at 06:57 PM.
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  2. #2
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    okay so i think i got it. my final answer was the matrix:
    -5/9 3 -7/3
    -83/27 2/3 25/3

    due to these weird fractions i get the feeling i made an error somewhere. anyone have any ideas??
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  3. #3
    MHF Contributor Calculus26's Avatar
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    the problem starts with your calculation of D(g)

    g(x,y)= (x^2 * 3y, 3x-y^2)

    so the matrix for D(g) is
    2xy^3 3x^2 y^2
    3 -2y
    Were you using g(x,y) = (x^2*y^3,3x-y2)

    or the formula for g(x,y) in your original post?

    I've included attachments for both --check my work I still disagree somewhat.
    Attached Thumbnails Attached Thumbnails quick question on multivariable calc chain rule problem-derivative.jpg   quick question on multivariable calc chain rule problem-derivative2.jpg  
    Last edited by Calculus26; September 28th 2009 at 11:33 PM.
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  4. #4
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    oh! it is y^3

    i'm sorry for the typo, don't worry about the problem, i found a mistake, fixed, it, and still got weird fractions in my answer, so i'm just going to hope it's right!
    Last edited by mr fantastic; September 30th 2009 at 03:17 AM. Reason: Merged posts
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  5. #5
    MHF Contributor Calculus26's Avatar
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    The attachment on the right was worked with y^3 do you get the same result?
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  6. #6
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    ah wow i actually got that!!! thank you for the reassurance
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