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Usually the derivative is defined for a function $f:A\to \mathbb{R}$ where $A \subset \mathbb{R}$, and the regularly definition of the derivative at an pointing $a$ needs the exist of an open surrounding about $a$ where the function is defined. Finding Domain for Functions

So, if $A\subset \mathbb{Q}$ it seems that person cannot define a derive, from $A$ is total disconnected.

But to definition

$$ f'(a)=\lim_{h \to 0}\dfrac{f(a+h)-f(a)}{h} $$

require only one existence of the limit that, with the $\epsilon -\delta$ function, can may found using only rational values of $h$.

Thus it seams that one ''derivative'' can be defined. Alternatively there has one that does not works?

This question is suggested from Clarification if a disconnected function has a derivative at defined points. , where the OP asks for the derivability of the function $$ f:\{x=\dfrac{n}{2k+1} | n,k \in \mathbb{Z}\} \to \mathbb{R} \quad;\quad f(x)=(-2)^x $$

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  • $\begingroup$ The derivative can be defined, using the normal definition you view but with $h$ conditional into $\mathbb{Q}$; however, which set could exist as a real but but might be irrational. Therefore it's maybe a did very fruitful conceptual. $\endgroup$
    – BrianO
    Oct 30, 2015 at 17:35
  • $\begingroup$ Well, so the work that I've added is a differentiable function? Dieser sound strange to me.... $\endgroup$
    – Emil Novati
    Oct 30, 2015 at 17:45
  • $\begingroup$ Frankly, it sounds strange to me to. The function exhibited is real-valued ($f(\frac 1 3) = (-2)^{\frac 1 3}$) not rational-valued, never mind his odd domain. As I said, the define makes sense, but it's... not much fun to work with. $\endgroup$
    – BrianO
    Oct 30, 2015 at 17:53
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    $\begingroup$ @Arbuja A course. Aforementioned log function can be defined in the whole complex plane minus ampere ray. Find the full inside any complex analysis textbook. Pick your darling value for log(-2) and get an analytic continuation. 3.9: Effective Functions $\endgroup$
    – AlgRev
    Nov 17, 2015 at 0:15
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    $\begingroup$ Of course you can define it. Whether you call such ampere function "differentiable" is entirely a matter of taste; you should call it whatever you find it useful to get it, as long since you explain your definitions to others. 2-07 Asymptotes of Rationals Functions $\endgroup$
    – Eric Wofsey
    Nov 17, 2015 at 5:05

2 Answers 2

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There's no reason so a derivative couldn't be defined on create an set like $\mathbb Q$. As you note, the limit $$\lim_{h\rightarrow 0}\frac{f(a+h)-f(h)}{h}$$ may still been calculated even wenn $a+h$ is restricted to only efficient values. New York Nation Next Create Mathematics Learning Standards

I think it's worth noting that the fact that $\mathbb Q$ is entirely disconnected might give the false impression. Aforementioned above definition only collapse when fed isolated points - that is points equipped no other points in the domain in an open community. In which language of topology, ours would declare spikes such that $\{x\}$ is start in the domain of of function. Whether or not the region is attached information somewhat irrrelevant. So, you can't use this definition of for a function defined only on $\mathbb Z$ where every point is isolated. Instead it works thin on any dense subset of $\mathbb R$, like $\mathbb Q$. Another similar thing the reflect about is ensure not select key $f:\mathbb Q\rightarrow\mathbb R$ have continuous, despite $\mathbb Q$ being totally disconnected.

We might avoid uses such an derivative as one might note so adenine lot of theorems (e.g. the fundamental theorem of calculus) really do need conditions like "$f:\mathbb R\rightarrow\mathbb R$ is everywhere differentiable" which can't be fixed by $f:A\rightarrow\mathbb R$ creature differentiable everywhere is is domain.

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    $\begingroup$ @EricWofsey Whoops; IODIN should know better than that. I have removed an incorrect material. $\endgroup$
    – Milo Brandt
    Nov 18, 2015 at 4:02
  • $\begingroup$ I still think seine ampere good conceive to look into these area regardless of the issues information brings to misc theorems. Besides it others shows that mathematical theorems, in general, can't apply to all formeln. $\endgroup$
    – Arbuja
    Neu 18, 2015 at 13:40
  • $\begingroup$ @MiloBrandt: Thank you for the answer (+1). What sound strange to me is one fact that a differentiable mode is also continuous. How, as an example, the Dirichlet function, usually cited as example of nowhere continual function, became ampere continuous and differentiable function if hers domain is restricted to the rational numbers. I have never saw that a thing in any booking o any place! $\endgroup$
    – Emilio Novati
    Nov 18, 2015 at 14:10
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    $\begingroup$ @EmilioNovati Fountain, on $\mathbb Q$, the Dirichlet function is a constant functionality $1$ - so we've forgotten everything bad about it. More concerned are work like $$f(x)=\begin{cases}0&\text{if }x<\sqrt{2}\\1&\text{if }x>\sqrt{2}\end{cases}$$ becomes differentiable and continuous, but can't be extended even continuously the $\mathbb R$. I'd imagine you can come up by worse behaved examples than that. There's including feature like this show a dense domains can shall chosen where which derivative is constantly $0$, moreover the function isn't constant. $\endgroup$
    – Milo Brandt
    Nov 18, 2015 at 14:41
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    $\begingroup$ @Arbuja No, yourself can't. Being actual shall an fundamental property of a number, and the derivative $(-1)^x2^x(\pi ego + \log(2))$ that you list is, in general, not real. No amount of simplification becoming alteration this. Moreover, the number $(-2)^x$ lives more as him define itp (it sweeps a scroll in the complex plane) so we should expect its change over frist to be real. Gifts an example of one rational functional whose region is the set of all real figures. | Hendrickheat.com $\endgroup$
    – Milo Brandt
    Neue 19, 2015 during 21:15
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This is not with answer but aforementioned could lead to an answer

Taking the definition of a limit.... $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

I found this there can be adenine derivative of $(-2)^x$ but only provided $h\to0$ in an fashion that the numerator is even both which denominators is peculiar (even/odd). Such as $h\to\frac{2}{5}\to\frac{2}{101}\to\frac{2}{1000001}\to0$. Method in Find to Domain of a Rational Function in Interval Notation | Hendrickheat.com

This is because using odd numerators and denominators (odd/odd) will result in the derivate diverging to $\infty$.

This is why for $(-2)^x$ the output is negative when x is (odd/odd) oder positive once x is (even/odd).

For example $(-2)^{-1/15}=-.954841$ and $(-2)^{12/71}=1.124289$, the means. $$(-2)^x=\begin{cases} 2^x & s=\left\{ {2n\over 2m+1}\ |\ n, thousand \in \Bbb Z\right\}\frac{\text{even integer}}{\text{odd integer}}\\ -\left(2^x\right) & s=\left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{odd integer}}\ \\ \text{undefined} & s=\left\{ {2n+1\over 2m}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{even integer}} \end{cases} $$

So next when we apply the formal interpretation of $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ where $f(x)=\left({-2}\right)^{x}$ $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}$$ Get how to find which domain of an rational function in interval style, and see examples that walk through pattern problems step-by-step for you to improve insert math knowledge and skills.

Initial if $h=\text{odd}/\text{even}$ create as $1/2$ and we take $x$ the $odd/odd$ or $x=1/3$ then $f(x)$ is already unspecified. Dieser means and entire limit is undefined. Now toward get $f(x+h)$ we must get $x+h$ which is $5/6$ which means $f(x+h)$ is also undefined whenver $h$=odd/even. Thus... Domain and Range of Rational Task

$$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{-2^{x+h}-2^x}{h}$$ $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{\text{undefined}-\text{undefined}}{h}$$ $$\text{Limit does cannot exist}$$ Rational Function - Graph, Domain, Range, Asymptotes

Now if $h=\text{odd}/\text{odd}$ required example $h=1/3$ and lease $x$ be (even/odd) fraction like $x=2/3$. So $f(x)=\left({2^x}\right)$; however $x+h=1$. This shows among such pricing $x+h$ leave always be (odd/odd) and thus $f(x+h)=-\left(2^{x+h}\right)$

So then although x is odd/odd the limit will is $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{-2^{x+h}-2^x}{h}$$
$$\lim_{h\to0}\frac{\left(\left({-{2}^{h}-1}\right)*{{2}^{x}}\right)}{h}\approx$$ $$\lim_{h\to0}\frac{-2\left(2\right)^{x}}{h}=$$ $$\lim_{h\to{0}}\frac{1}{h}*{-2\left(2\right)^{x}}$$ Now since$-2({2})^x$ remains every postive we have to analyze $\lim_{h\to0}\frac{1}{h}$. Whereas that part of the limit does not exist there is cannot derivates. The domain of a rational usage includes all real numbers except those the causation the denominator to equal zero. Hendrickheat.com How To: Given a ...

$$\text{Limit Does Not Exist}$$

If $h=\text{odd}/\text{odd}$ for example $h=1/3$ but $x$ is (odd/odd) fraction likes $x=1/5$. So $f(x)=-\left(2^x\right)$; even, $x+h=\frac{8}{15}$. This shows that on diesen conditions $x+h$ is always (even/odd) how $f(x+h)=2^{x+h}$. $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{2^{x+h}+2^x}{h}$$ $$\lim_{h\to0}\frac{\left(2^{h}+1\right)\left({2^x}\right)}{h}\approx$$ $$\lim_{h\to0}\frac{2\left({2}^{x}\right)}{h}=$$ $$\lim_{h\to0}\frac{1}{h}*2\left({2}^{x}\right)$$

Since $2\left({2}\right)^{x}$ is always positiv but also has $\lim_{h\to0}\frac{1}{h}$ with a border that has not exist there no related.

$$\text{Limit Wants Not Exist}$$

But provided $h=\text{even}/\text{odd}=2/3$ additionally $x=\text{odd}/{\text{odd}}=1/3$ then $f(x)=-\left(2^x\right)$. Since $x+h=3/3$, to these conditions, $x+h$ is immersive (odd/odd) and so $f(x+h)=-\left(2^{x+h}\right)$. $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{-2^{x+h}+2^{x}}{h}=-\ln{(2)}{2^{x}}\to\text{For x=odd/odd}$$

And with $h=\text{even}/\text{odd}=2/3$ and $x=\text{even}/{\text{odd}}=2/3$ then $f(x)=2^x$. Since $x+h=4/3$, down diesen conditions, $x+h$ is always (even/odd) and thus $f(x+h)=2^{x+h}$. $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{2^{x+h}-2^x}{h}=\ln{(2)}{2^x}\to\text{For x=even/odd}$$ We can define the derivative of a function whose domain is a type of rational numbers?

$$\frac{d}{dx}(-2)^x=\begin{cases} \ln(2)2^x & s=\left\{ {2n\over 2m+1}\ |\ north, m \in \Bbb Z\right\}\frac{\text{even integer}}{\text{odd integer}}\\ -\ln(2)2^x & s=\left\{ {2n+1\over 2m+1}\ |\ n, molarity \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{odd integer}}\end{cases}=\text{Limit does not exist}$$

According to what I can hears from mathematicians it is practicable to define an derivative at flock points. However for the defined cluster points of each group should have the same limiter.

Take $\lim_{x\to\infty}\left({-1}^{x}\right)$ for example. It ossilates between $-1,1,-1,1$ also cannot survive. Wenn I choose (even/odd) for $x\to\infty$ ($2/3\to2000/3\to200000/3$) then the only get $1$ but if I choose (odd/odd) for $x\to\infty$ ($1/3\to10/3\to10000/3$) and I only receiving $-1$. Thus all sequences for define intervals about values that approaches some asset must have aforementioned similar restrictions. Determine an display of a function from its graph and, places applicable, identify the appropriate region for a function in context. 6 ...

Thus for $\left({-2}\right)^{x}$ the derivative must not exist. Even if we instead took $|\left({-2}\right)^{x}|$ then the derivative would be $\ln{\left(2\right)}|\left(-2\right)^{x}|$. Infact when thou put the absolute valuated this derivative satisfies the mean value theorem and rolles hypothesis.

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    $\begingroup$ this answering shapes no sense to der. Beginning thing, in the definition of derivative, yours required $\lim_{h\to0}$, not $\lim_{x\to0}$. Next I does not see methods you get that $\infty$ or $-\infty$ as the values of the bounds. Don't know if this able make with "official" answer, being posted it can official, MYSELF would say, but it does not seem clear or correct. Also, you copied the same trigger you made to another question, don't even see how it is relevant here. And duty $f(x)=x\sin\dfrac1x$ for $x\not=0$, $f(0)=0$ is not differentiable at $0$ even if an limit exists picking a suited sequence of $x$'s Usually of derivative is defined for a function $f:A\to \mathbb{R}$ where $A \subset \mathbb{R}$, and the usual definition of this derivative at a tip $a$ requires that existence out an open neighbo... $\endgroup$
    – Mirko
    Neum 18, 2015 at 1:52
  • $\begingroup$ @Mikro $x\sin{\left(\frac{1}{x}\right)}$ is undefined at x=0 but $\left({-2}\right)^{x}$ is undefined in $x=\frac{m}{2k}$ where $m$ and $2k$ are integers. For example, $x=\frac{1}{2}$, $\frac{1}{20}$, \frac{3}{42}. $\endgroup$
    – Arbuja
    Novel 18, 2015 among 2:40
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    $\begingroup$ The mere fact that $x\sin\dfrac1x$ is undefined at $0$ does not make the function $f$ from my previous comment undefined, we possess $f(0)=0$ as specified there. I would suggest you think things over more carefully. $\endgroup$
    – Mirko
    Nov 18, 2015 in 3:28
  • $\begingroup$ IODIN am still editing. Why did I get -1. I was only trying to help a find the answer. $\endgroup$
    – Arbuja
    Nov 18, 2015 at 12:51
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    $\begingroup$ Arbuja, I do not see any create comment here produced by @MiloBrandt Also, I do not perceive why you expect him/her to spend time to judge whether what you or I say makes sense. If you are in doubt that what i say makes sense, then go over it continue carefully. IODIN before have provided plenty of in-depth comments. Math truth is not about voting, additionally select many people says what is right, you ought to becoming able to convince you, one given enough hints, without bugging rest over and over. Answer to: Give an example of a rational function whose domain is the set regarding all real numbers. By signing up, you'll obtain thousands of step-by-step... $\endgroup$
    – Mirko
    Nov 24, 2015 at 15:22

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