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Partial derivatives

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Let Z = 3x-2y x = u+v ln(u) and y = u^2-v ln(v)

    2. Relevant equations

    find Pdz/Pdu and Pdz/Pdv Pd (partial Derivative)

    3. The attempt at a solution

    Pdz/Pdx = 3 Pdz/Pdy = -2

    Pdx/Pdu = v/u +1 Pdx/Pdv = ln(u)

    Pdy/Pdu = 2u Pdy/Pdv = -ln(v)-1




    Pdz/Pdu = Pdz/Pdx * Pdx/Pdu + Pdz/Pdy * Pdy/Pdu

    = 3 * (v/u +1) + -2 * 2u

    =3(u+v)/u + (-4u)

    = (-4u^2 - 3u - 3v)/u

    Could someone please let me know if this is correct?
    regards
    Brendan
     
  2. jcsd
  3. Mar 30, 2009 #2

    tiny-tim

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    Hi Brendan! :smile:

    (have a curly d: ∂ and try using the X2 tag just above the Reply box :wink:)
    Yes that's fine! :smile:

    (except for the minuses in the last line :wink:)
     
  4. Mar 30, 2009 #3
    Thanks alot for your help.
    regards
    Brendan
    P.S what does the X^2 tag do?
     
  5. Mar 30, 2009 #4

    tiny-tim

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    If you click the tag, it types [noparse], and anything in between goes above the line and gets smaller …

    so you see yx2n in the Reply box, but in the post you see[/noparse] yx2n :smile:
     
  6. Mar 30, 2009 #5
    For the same equation,

    Let Z = 3x-2y x = u+v ln(u) and y = u^2-v ln(v)

    To find ∂z/∂v

    ∂z/∂v = ∂z/∂x * ∂x/∂v + ∂z/∂y * ∂y/∂v

    We have ∂z/∂x = 3, ∂x/∂v = ln(u), ∂z/∂y= -2, and ∂y/∂v= -ln(v)-1

    Is it:

    3*ln(u)-2*(ln(v)+1) ?

    regards
    Brendan
     
  7. Mar 30, 2009 #6

    tiny-tim

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    Yes that's fine! :smile:

    (except for the minuses in the last line … again :rolleyes:)
     
  8. Mar 30, 2009 #7
    Thanks once again!
     
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