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Implicit differentiation problem.

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Find y´(x) for x + y = 1/x^2 + 8/y^2


    2. Relevant equations



    3. The attempt at a solution
    Rewrite the eq. as
    x + y = x^-2 + 8y^-2
    differentiate.
    1 + y´(x) = (-2x^-3) + (-16y^-3)(y´(x))
    Rearrange
    1 + (2x^-3) = (-16y^-3)(y´(x)) - (y´(x))
    (1 + (2x^-3))/(16y^-3) = (-2y´(x))
    (1 + (2x^-3))/((-2)16y^-3) = y´(x)

    Now I know that the answer should look like this:
    (1+2x^−3 )/(−1−16y^−3).

    What am i doing wrong?
     
  2. jcsd
  3. Oct 20, 2009 #2
    Something looks really strange for the 2nd line. Careless mistake?
     
  4. Oct 20, 2009 #3
    How do i not eliminate the y´(x) term? Im probably doing some really basic mistake that should be enough motivation to make me drop calculus.

    1 + (2x^-3) = (-16y^-3)(y´(x)) - (y´(x))
    (1 + (2x^-3))/(-16y^-3) = (y´(x)) - (y´(x))
    (1 + (2x^-3))/(-16y^-3) = 0
     
  5. Oct 21, 2009 #4
    You want to make y'(x) the subject, hence you factor it out.
    1 + (2x^-3) = (-16y^-3)(y´(x)) - (y´(x)) = y'(x) (-16y^-3 - 1)
    This appears to be an algebra issue that you are facing, so you might want to work on it.
     
  6. Oct 21, 2009 #5
    From: 1 + (2x^-3) = (-16y^-3)(y´(x)) - (y´(x))
    You cannot divide by (-16y^-3) on both sides, in order to get your next expression, because the last part of your equation -> " - y´(x) " doesn't have that factor in it.

    Instead, "pull out" y´(x) from the second half of the equation giving:

    1 + (2x^-3) = y´(x)*[-16y^-3 - 1] (I put in brackets to make it a bit easier.

    From here it should be pretty easy, just divide by " -16y^-3 - 1"
     
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