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Differntiating natural logs of x and y functions

  1. Mar 22, 2006 #1
    i have a question I'm attempting as extra work, I have tried the usual method but no luck.... find expressionsf for dy/dx in terms of x and y:

    ln(x^2+1) + ln(y+1) = x +y
     
  2. jcsd
  3. Mar 22, 2006 #2
    Do you know how to do implicit differentiation?
     
  4. Mar 22, 2006 #3
    yes, thats the method i tried, but i get the wrong answer
     
  5. Mar 22, 2006 #4

    Hootenanny

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    If you show your working we'll try to correct you.
     
  6. Mar 22, 2006 #5
    dy/dx - 1/ x^2 +1 + 1 / y+1.dy/dx = 1 + dy/dx

    =1 /x^2+1 - 1 = dy/dx(1-1/y+1)

    dy/dx = (x^2 + 1)^-1 - 1/ 1- (y+1)^-1
     
  7. Mar 22, 2006 #6
    Remember [tex] \frac{d}{dx} \ln(x^2 +1) \neq \frac{1}{x^2+1} [/tex]

    You need to use the chain rule to get the proper derivative.
     
    Last edited: Mar 22, 2006
  8. Mar 22, 2006 #7

    Hootenanny

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    Yes you need to multiply this by the derivative of the bracket :smile:
     
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