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Derivative of natural log

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Find y'

    y=ln(x^2 + y^2)

    2. Relevant equations
    d/dx ln(u)= 1/u du/dx

    3. The attempt at a solution

    y' = [1/(x^2 + y^2)] (2x + 2y)
    y' = (2x+2y)/(x^2 + y^2)

    But my book says the answer is 2x/(x^2 + y^2 - 2y)

    How can that be?
  2. jcsd
  3. Oct 16, 2007 #2


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    Homework Helper

    You need to be differentiating the y on the right hand side of the equation when you take the derivative of the entire equation.
  4. Oct 16, 2007 #3
    Are you saying my first step should be:
    dy/dx = (1/x^2 + y^2) [2x+2y(dy/dx)]
  5. Oct 16, 2007 #4

    this problem must be solved by implicit differentiation
  6. Oct 16, 2007 #5
    I don't understand what the 2nd step would be though. So I have dy/dx on both sides, but isn't dy/dx what I want to solve for? Wouldn't they cancel each other out?
  7. Oct 17, 2007 #6


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    No, they won't cancel because one is multiplied by [itex][2y/(x^2+ y^2)][/itex]
    (Please, don't write [itex]1/x^2+ y^2[/itex]! That's a completely different value).

    Solve your equation for dy/dx.
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