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Calculus partial derivatives problem [y^(-3/2)arctan(x/y)] * *

  1. Oct 11, 2012 #1
    Calculus partial derivatives problem [y^(-3/2)arctan(x/y)] *urgent*

    1. The problem statement, all variables and given/known data

    f(x,y) = y^(-3/2)arctan(x/y)...find fx(x,y) and fy(x,y) [as in derivatives with respect to x and with respect to y].

    2. Relevant equations



    3. The attempt at a solution

    mathematics is not my strong suit..i tried the problem from a couple of different angles..i am not getting the correct answer


    here is what i have tried doing so far..i used the product rule obviously. we know that derivative of (1/a)arctan(x/a) gives (1/(a^2+x^2)

    i took y^(-3/2) as 'u' and arctan(x/y) as 'v' for the implementation of the product rule. so am i getting y/(y^2+x^2) as the derivative (since (1/y) is missing from the arctan term)?

    for fx(x,y) i get the answer y^(-1/2)/(y^2+x^2). this answer matches with the book's answer (ch-13.3 prob no.25 - calculus 9th ed by anton, bivens, davis)

    however i am not sure if i got it correct only by chance since i used the same method for fy(x,y) only to get an incorrect answer..my answer for fy(x,y) came
    -(3/2)y^(-5/2)arctan(x/y) - xy^(-5/2) / (y^2+x^2)

    the correct answer is
    -(3/2)y^(-3/2)arctan(x/y) - xy^(-3/2) / (y^2+x^2) [note: (xy^(-3/2)...not (xy^(-5/2)]

    fy(x,y) = -(3/2)(y^(-5/2))arctan(x/y) + (y^(-3/2)) (y/(y^2+x^2)) (-x/(y^2))
    which gives: -(3/2)y^(-5/2)arctan(x/y) - (xy^(-5/2)) / (y^2+x^2)
    so what went wrong there..?

    been stuck for hours its quite frustrating....so can anyone please show me the workings with the steps so that i know where i am getting it wrong?

    thanks in advance! :)
     
  2. jcsd
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