# Derivative of Ln(x)

1. Feb 20, 2011

### SprucerMoose

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. Feb 20, 2011

### dextercioby

Yes to all your questions. Rigorously speaking, the domain of defintion for the functions being integrated or differentiated must always be specified.

3. Feb 20, 2011

### 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. $image=http://latex.codecogs.com/gif.latex?\int_{-1}^{1}(ln|x|+5)+dx&hash=1169e239e9f0ba4b1f6894f87f2610e9$

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.

4. Feb 20, 2011

### Redbelly98

Staff Emeritus
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.

5. Feb 20, 2011

### SprucerMoose

Oops, I meant $image=http://latex.codecogs.com/gif.latex?\int_{-1}^{1}ln%28|x|+5%29%20dx&hash=c3f17d0d0e4b5e3899c326030f3f6c2b$

6. Feb 21, 2011

### PhDorBust

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.

7. Feb 21, 2011

### Redbelly98

Staff Emeritus
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