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Derivative of a function with ln

  1. Oct 16, 2007 #1
    This is a small part of a bigger problem, but the part I am having trouble with is finding the derivative of...
    I'm sure it is something simple and I remember learning it in Calc I or II but I forgot. Please help remind me! Thank you!!
  2. jcsd
  3. Oct 16, 2007 #2


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    [tex]ln(\sqrt{x^2+y^2}) = ln(\sqrt{x^2+y^2})^\frac{1}{2} = \frac{1}{2}ln(x^2+y^2)[/tex]

    Is it easier now?
  4. Oct 16, 2007 #3
    Yes I believe so...

    Would it then be...

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

    ...if I remember correctly??
  5. Oct 16, 2007 #4
    Which would then simplify to


  6. Oct 16, 2007 #5
    Are you taking the total derivative, or what is the variable you are differentiating with respect to?
  7. Oct 16, 2007 #6


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    Remember [tex]\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}[/tex]
  8. Oct 17, 2007 #7


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    Supraanimo's question is still important. ln[sqrt(x^2+y^2)] has two variables. "The derivative" might be the gradient, but more often you are asked to find the partial derivatives. Which is it?
  9. Oct 17, 2007 #8


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    Just to correct a typo...This should read

    [tex]ln(\sqrt{x^2+y^2}) = ln((x^2+y^2)^\frac{1}{2}) = \frac{1}{2}ln(x^2+y^2)[/tex]
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