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Derivative of Ln(x)

  1. Feb 20, 2011 #1
    Hey guys,
    I just have a quick question about the derivative of ln(x). If i was to calculate the derivative of ln(x +1) = 1/(x+1), would I technically have to restrict the domain of the solution to x>-1?
    Otherwise when I take the antiderivative again, I will have Ln|x+1| (note the absolute value) and not the original function.

    One other question that I sort of implied to be fact, if i take the integral of 1/x, where x>0, is the solution now Ln(x) rather than Ln|x|?
  2. jcsd
  3. Feb 20, 2011 #2


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    Yes to all your questions. Rigorously speaking, the domain of defintion for the functions being integrated or differentiated must always be specified.
  4. Feb 20, 2011 #3
    Thanks for the quick response.

    By the way can a definite integral be calculated across an interval where the function is continuous, but not differentiable for the entire interval?

    ie. http://latex.codecogs.com/gif.latex?\int_{-1}^{1}(ln|x|+5) dx [Broken]

    I'm not talking about splitting the interval to calculate the area, if that is what would be done, I'm more interested in the specific rules for a definte integral.
    Last edited by a moderator: May 5, 2017
  5. Feb 20, 2011 #4


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    The function ln|x|+5 is not continuous in the interval [-1,1], so it is not illustrative of your question. There is a discontinuity at x=0.

    I'm pretty sure every function is integrable over any range in which it is continuous. A better example, for you question, would be to integrate y=x1/3 over some range that includes x=0, since this function is continuous but not differentiable at x=0.
  6. Feb 20, 2011 #5
    Oops, I meant gif.gif
  7. Feb 21, 2011 #6
    Sure, when you take a course in real analysis, you'll learn that continuity on [a,b] implies that the definite integral on [a,b] exists.

    A simple example would be [tex]\int_{-1}^1 | x| \, dx[/tex] which is continuous but not differentiable at 0. A more satisfying example would be the Weierstrass function which is continuous but nowhere differentiable.
    Last edited by a moderator: May 5, 2017
  8. Feb 21, 2011 #7


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    Okay, understood.

    The only rule I can think of is: whenever the argument of an absolute value function changes sign, you must split the integration interval at that (those) point(s). We can't come up with an analytic form for integrating |f(x)| without doing that.
    Last edited: Feb 21, 2011
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