Derivative of Ln(x): Domain Restriction & Antiderivative

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In summary: 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.
  • #1
SprucerMoose
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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|?
 
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  • #2
Yes to all your questions. Rigorously speaking, the domain of defintion for the functions being integrated or differentiated must always be specified.
 
  • #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

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.
 
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  • #4
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
Oops, I meant
gif.gif
 
  • #6
SprucerMoose said:
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

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 [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.
 
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  • #7
SprucerMoose said:
Oops, I meant
gif.gif
Okay, understood.

SprucerMoose said:
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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FAQ: Derivative of Ln(x): Domain Restriction & Antiderivative

What is the domain restriction for the derivative of Ln(x)?

The domain restriction for the derivative of Ln(x) is x > 0. This is because the natural logarithm function Ln(x) is only defined for positive real numbers. Therefore, the derivative of Ln(x) is also only defined for positive real numbers as well.

Why is the domain restricted for the derivative of Ln(x)?

The domain is restricted for the derivative of Ln(x) because the natural logarithm function is not defined for negative or zero values. The derivative represents the instantaneous rate of change at a specific point on a curve, and in order for this to be defined, the original function must also be defined at that point.

Is the domain restriction for the derivative of Ln(x) inclusive or exclusive?

The domain restriction for the derivative of Ln(x) is exclusive, meaning that the value of x must be strictly greater than 0. This is because the natural logarithm function is not defined at 0, and the derivative is undefined at that point as well.

What is the antiderivative of Ln(x)?

The antiderivative, or the inverse function, of Ln(x) is e^x. This means that e^x is the function whose derivative is equal to Ln(x). In other words, if we take the derivative of e^x, we will get Ln(x) as the result.

Can the antiderivative of Ln(x) be expressed in terms of elementary functions?

Yes, the antiderivative of Ln(x) can be expressed in terms of elementary functions, specifically the exponential function e^x. However, it cannot be expressed as a finite combination of elementary functions, meaning it cannot be written using only basic algebraic operations, exponential and logarithmic functions, and trigonometric functions.

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