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For instance, The function $f(x)=\ln (x) $ is defined for $x>0$, but $f'(x)=\dfrac{1}{x}$ is defined over the interval $(-\infty ,0)\cup (0,\infty )$. Are there any other functions $f$ other than the natural logarithm function that have this property?
Does this relate to the integral of $\dfrac{1}{x}$ being $\ln |x|$ by any means? How would you explain that for a student that just finished calculus? I mean it's okay to use higher level mathematics but I would most probably be confused, as I haven't yet got into analysis nor did I take any proofs classes.

commie trivial
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Dreamer
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  • How would you define such a derivative? – Xander Henderson Jul 24 '22 at 16:03
  • Do you mean like, a derivative means slope of tangent to the function? – Dreamer Jul 24 '22 at 16:10
  • Actually on the negative side of $\mathbb{R}$, $\ln(-x)$ is defined and its derivative is $1/x$. So yes, $\ln|x|$ is the answer. – Jean-Armand Moroni Jul 24 '22 at 16:12
  • If you mean that the derivative is the slope of a line tangent to a function at a particular point, then if $f$ is not defined at $a$, how would you define a tangent line at $a$? Also, I doubt that this is the actual definition of the derivative that you were given. Typically, the derivative is defined in terms of a limit. Look at that definition. How would you define that limit at a point $a$ if $a$ is not in the domain of your function? – Xander Henderson Jul 24 '22 at 16:15
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    Does this mean that $f'(x)= \dfrac{1} {x}$ is only defined for $x>0$? – Dreamer Jul 24 '22 at 16:19
  • In the particular case you give, the function $x \mapsto 1/x$ is defined for all nonzero real numbers, and is integrable over any interval which does not contain zero. It's integral is given by $$ \int_{a}^{b} \frac{1}{x} \mathrm{d}x = \begin{cases} \log(b)-\log(a) & \text{if $a,b > 0$, and} \ \log(-a)-\log(-b) & \text{if $a,b < 0$.} \end{cases} $$ This can be abbreviated to $\log|x|$, which is differentiable everywhere except zero (where it is not defined). – Xander Henderson Jul 24 '22 at 16:20
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    The statement $f'(x) = 1/x$ is a differential equation. It has many solutions, which can be summarized (more or less) as $$ f(x) = \begin{cases} \log(x) + C & \text{if $x > 0$, and} \ \log(-x) + D & \text{if $x < 0$,} \end{cases} $$ where $C$ and $D$ are constants. – Xander Henderson Jul 24 '22 at 16:22
  • So $f(x)= ln(x)$ is just a "special case" of $\int_{a}^{b} \frac{1}{x} \mathrm{d}x$ where $x>0$ and C = 0? – Dreamer Jul 24 '22 at 16:25
  • It is not clear to me what problem you are trying to solve, hence I don't really know how to answer your question, but... yes? – Xander Henderson Jul 24 '22 at 16:26
  • Nevermind, I think that answers my question, thank you very much! – Dreamer Jul 24 '22 at 16:30
  • @Dreamer This answer may be of interest. BTW, contrary to your first sentence, $f(x)=\ln(x)$ implies that $f'$ has domain $(0,\infty).$ – ryang Jul 24 '22 at 16:42

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Given a function $f:X\to Y$ where $X\subseteq\mathbb{R}$ and $Y\subseteq\mathbb{R}$ and $a\in X$ is also an accumulation point of $X$, we define the derivative of $f$ at $a$ as the limit (when it exists): \begin{align*} f'(x) = \lim_{x\to a}\frac{f(x) - f(a)}{x - a} \end{align*}

As you can see, the definition of limit demands that $a$ belongs to the domain of $f$.

Having said that, the answer to your question is: no.

Átila Correia
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  • Yes, I guess what he/she wants to ask is after taking the derivative on the domain where it is originally defined, then the new function (forget where it originates from) has a larger domain? – MathFail Jul 24 '22 at 20:49