# Derivative of Ln(x)

by SprucerMoose
Tags: derivative
 P: 67 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|?
 Sci Advisor HW Helper P: 11,927 Yes to all your questions. Rigorously speaking, the domain of defintion for the functions being integrated or differentiated must always be specified.
 P: 67 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. $\int_{-1}^{1}(ln|x|+5) dx$ 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.
 Mentor P: 12,070 Derivative of Ln(x) 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.
 P: 67 Oops, I meant $\int_{-1}^{1}ln(|x|+5) dx$
P: 142
 Quote by SprucerMoose 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. $\int_{-1}^{1}(ln|x|+5) dx$ 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.
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 $$\int_{-1}^1 | x| \, dx$$ which is continuous but not differentiable at 0. A more satisfying example would be the Weierstrass function which is continuous but nowhere differentiable.
Mentor
P: 12,070
 Quote by SprucerMoose Oops, I meant $\int_{-1}^{1}ln(|x|+5) dx$
Okay, understood.

 Quote by SprucerMoose 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.
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.

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