Derivative of Ln(x)

  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. dextercioby

    dextercioby 12,327
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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. 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. [​IMG]

    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.
  5. Redbelly98

    Redbelly98 12,046
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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. Oops, I meant [​IMG]
  7. 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.
  8. Redbelly98

    Redbelly98 12,046
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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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