# I Two questions about derivatives

#### Math_QED

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In Rudin, the derivative of a function $f: [a,b] \to \mathbb{R}$ is defined as:

Let $f$ be defined (and real-valued) on $[a,b]$. For any $x \in [a,b]$, form the quotient $\phi(t) = \frac{f(t) - f(x)}{t-x}\quad (a < t <b, t \neq x)$ and define $f'(x) = \lim_{t \to x} \phi(t)$, if the limit exists.

An analoguous definition is given for functions defined on an open interval.

Q1: Why do we restrict ourselves to defining derivatives on intervals? I would propose the following definition:

Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function and $x$ a limit point of $E$.

Define $$\phi: E \setminus \{x\} \to \mathbb{R}: t \mapsto \frac{f(t) - f(x)}{t-x}$$. Then we define $$f'(x) = \lim_{t \to x} \phi(t)$$ if the limit exists.

Q2: The quotient rule in Rudin is written in the following way:

Suppose $f,g$ are real functions on $[a,b]$ that are differentiable at $x \in [a,b]$. Then $f/g$ is differentiable at $x$, provided that $g(x) \neq 0$ and

$\left(\frac{f}{g}\right)'(x) = \frac{g(x)f'(x) - g'(x)f(x)}{g^2(x)}$

Following Rudin's convention, the domain of $f/g$ are those point $x$ of $[a,b]$ where $g(x) \neq 0$.

However, Rudin's definition of differentiability only mentions the cases where a function is defined on an open or closed interval, while $f/g$ may be defined on a domain that it is not an interval.

How should this be interpreted?

I see two solutions:

(1) We can restrict the function to an interval where it is non-zero (but the theorem does not mention this!)
(2) We use the alternative definition I gave in Q1.

Any thoughts?

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#### FactChecker

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What would you use your definition for? There are all sorts of possibilities in mathematics that do not have enough use to be considered. Your suggestion is fine if there is a reason for it.

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#### Math_QED

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What would you use your definition for? There are all sorts of possibilities on mathematics that do not have enough use to be considered. Your suggestion is fine if there is a reason for it.
For example, to resolve Q2, and to have a general definition that works for all sets.

#### FactChecker

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If you define a derivative of a quotient at a point where it is undefined, how would you use that? What you consider a "problem" can only be a problem if it stops you from doing something that needs to be done.

#### Math_QED

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If you define a derivative of a quotient at a point where it is undefined, how would you use that? What you consider a "problem" can only be a problem if there it stops you from doing something that needs to be done.
Good point. Will think about it some more.

#### FactChecker

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It's good that you are thinking in abstract generalities like that. If you are working on something and your suggestion helps, then do it. Till then, it has a drawback that it restricts what you can say about a function with a derivative. (It no longer implies continuity; It complicates the fundamental theorems of calculus.) So it may not be good for general use.

#### Math_QED

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It's good that you are thinking in abstract generalities like that. If you are working on something and your suggestion helps, then do it. Till then, it has a drawback that it restricts what you can say about a function with a derivative. (It no longer implies continuity; It complicates the fundamental theorems of calculus.) So it may not be good for general use.
Thanks. I guess that answers question 1 for now. Maybe you have an idea about question two too?

#### FactChecker

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I see two solutions:

(1) We can restrict the function to an interval where it is non-zero (but the theorem does not mention this!)
You mean where the denominator is nonzero, right?
(2) We use the alternative definition I gave in Q1.
That definition probably needs more thought. What does it say about points that are not a cluster point of E? Do they automatically have a derivative? If so, what is the value?

#### Math_QED

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You mean where the denominator is nonzero, right?That definition probably needs more thought. What does it say about points that are not a cluster point of E? Do they automatically have a derivative? If so, what is the value?
Yes, where the denominator is non-zero.

I wish not to define limits for isolated points, because we won't have unicity of limits anymore.

#### FactChecker

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Yes, where the denominator is non-zero.

I wish not to define limits for isolated points, because we won't have unicity of limits anymore.
What you are suggesting may be possible with more work. In fact, it may have been done and I am just not familiar with it. I'll leave further comments for others who know more about it.

#### mathman

For question 2. Rudin's definition for the derivative of a quotient f/g asserts that f and g are differentiable at that point. This implies some interval around that point for f and g..

#### Math_QED

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For question 2. Rudin's definition for the derivative of a quotient f/g asserts that f and g are differentiable at that point. This implies some interval around that point for f and g..
Yes, and then we restrict f/g to this interval. My concern was that Rudin doesn't mention this.

Thanks, now I see that my concern was valid.

#### Math_QED

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It's good that you are thinking in abstract generalities like that. If you are working on something and your suggestion helps, then do it. Till then, it has a drawback that it restricts what you can say about a function with a derivative. (It no longer implies continuity; It complicates the fundamental theorems of calculus.) So it may not be good for general use.
We can undo your drawbacks by adding another assumption, namely that the limit point must be a point of the set itself. Then differentiability implies continuity and the fundamental theorem of calculus is just this definition applied to an interval.

#### FactChecker

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We can undo your drawbacks by adding another assumption, namely that the limit point must be a point of the set itself. Then differentiability implies continuity and the fundamental theorem of calculus is just this definition applied to an interval.
Ok. I'll buy that. I just have trouble visualizing all the possible consequences. I guess the charactoristic function of the rational numbers would have the newly defined "derivative" on the rationals and on the irrationals but would not be continuous anywhere in the real line. It's something I would not want to put a lot of work into unless I thought it would lead to something profound.

#### Stephen Tashi

Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function and $x$ a limit point of $E$.

Define $$\phi: E \setminus \{x\} \to \mathbb{R}: t \mapsto \frac{f(t) - f(x)}{t-x}$$.
Your notation assumes $f$ has a defined value at $x$. You may as well make this assumption explicit by saying that $E$ is a closed set.

However, Rudin's definition of differentiability only mentions the cases where a function is defined on an open or closed interval, while $f/g$ may be defined on a domain that it is not an interval.
The simplest solution is interpret the statement "$f$is defined on $[a,b]$ " to mean that the domain of $f$ is at least $[a,b]$ but may be greater.

For example, Let $h(x) = f(x)/g(x)$ and let $g(c) = 0$. If there is an interval $[a,b], b < c$ where both $f(x)$ and $g(x)$ are defined then the derivative of $h$ can be defined for each $x$ in $[a,b]$

It's possible that Rudin intends such an interpretation. There is a tradition in mathematical writing that statements like "N has a divisor" do not exclude possibilities like N having 30 different divisors. But that tradition, if we want to say $f$ is defined on a set $E$ and nowhere else, we would say $f$ is defined exactly on $E$.

Perhaps that's why Rudin says "$f$ is defined on $[a,b]$ instead of saying "The domain of $f$ is $[a,b]$.

#### mathwonk

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in general a derivative is a linear map that aproximates locally a given non linear map. in order for this to make sense, the domain of the derivative should approximate locally the domain of the non linear map. now the domain of a linear function is a vector space, so the non linear function should have as a domain some space that is locally approximated by a vector space, namely the domain of the non linear function should be a curved space made out of a neighborhood of the origin of a vector space, i.e. it should be a "manifold", or in dimension one, an open interval.

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#### Math_QED

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Your notation assumes $f$ has a defined value at $x$. You may as well make this assumption explicit by saying that $E$ is a closed set.

The simplest solution is interpret the statement "$f$is defined on $[a,b]$ " to mean that the domain of $f$ is at least $[a,b]$ but may be greater.

For example, Let $h(x) = f(x)/g(x)$ and let $g(c) = 0$. If there is an interval $[a,b], b < c$ where both $f(x)$ and $g(x)$ are defined then the derivative of $h$ can be defined for each $x$ in $[a,b]$

It's possible that Rudin intends such an interpretation. There is a tradition in mathematical writing that statements like "N has a divisor" do not exclude possibilities like N having 30 different divisors. But that tradition, if we want to say $f$ is defined on a set $E$ and nowhere else, we would say $f$ is defined exactly on $E$.

Perhaps that's why Rudin says "$f$ is defined on $[a,b]$ instead of saying "The domain of $f$ is $[a,b]$.
Thanks for your insight. Asking for a closed set doesn't seem to make much sense though, as it also seems common to define derivatives on an open interval.

Another advantage of my definition would be that we can consider derivatives of a function like $f: \mathbb{R}_0 \to \mathbb{R} : x \mapsto \frac{1}{x}$ without having to split up the domain in two open intervals or something like that.

#### Math_QED

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Ok. I'll buy that. I just have trouble visualizing all the possible consequences. I guess the charactoristic function of the rational numbers would have the newly defined "derivative" on the rationals and on the irrationals but would not be continuous anywhere in the real line. It's something I would not want to put a lot of work into unless I thought it would lead to something profound.
What do you mean with characteristic function? The function that is 1 on the rationals and 0 elsewhere? If you put $E = \mathbb{Q}$, then this function is continuous on $E$ (it is constant there) with derivative 0. Same story if you put $E = \mathbb{R} \setminus \mathbb{Q}$.

That we can interpret such a derivative as the direction of a tangent line or that we can give it physical meaning are other questions.

#### FactChecker

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What do you mean with characteristic function? The function that is 1 on the rationals and 0 elsewhere?
Yes
If you put $E = \mathbb{Q}$, then this function is continuous on $E$ (it is constant there) with derivative 0. Same story if you put $E = \mathbb{R} \setminus \mathbb{Q}$. That we can interpret such a derivative as the direction of a tangent line or that we can give it physical meaning are other questions.
I agree with all that. It's good that you realize the possibilities of making the definition more general.

#### lavinia

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Q1: Why do we restrict ourselves to defining derivatives on intervals? I would propose the following definition:

Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function and $x$ a limit point of $E$.

Define $$\phi: E \setminus \{x\} \to \mathbb{R}: t \mapsto \frac{f(t) - f(x)}{t-x}$$. Then we define $$f'(x) = \lim_{t \to x} \phi(t)$$ if the limit exists.
The geometric idea of differentiability is that there is a tangent line to the graph of $f$. The derivative is the slope of the tangent line. If the function does not have a tangent line, then it is not thought of as differentiable.

Let $E$ be the rational numbers . Let $f(x) = 0$ on $E$ and $f(\sqrt 2)=0$. For every other irrational number, let $f(x)=x$.

On $E$ the limit of the Newton quotient is $0$ at $\sqrt 2$ but along a sequence of irrationals the Newton quotient is $(x+h-0)/h$ and this diverges.

In this case there is no tangent line at $\sqrt 2$ even though the limit of the Newton quotient exists along $E$.

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#### Math_QED

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The geometric idea of differentiability is that there is a tangent line to the graph of $f$. The derivative is the slope of the tangent line. If the function does not have a tangent line, then it is not thought of as differentiable.

Let $E$ be the rational numbers . Let $f(x) = 0$ on $E$ and $f(\sqrt 2)=0$. For every other irrational number, let $f(x)=x$.

On $E$ the limit of the Newton quotient is $0$ at $\sqrt 2$ but along a sequence of irrationals the Newton quotient is $(x+h-0)/h$ and this diverges.

In this case there is no tangent line at $\sqrt 2$ even though the limit of the Newton quotient exists along the rational numbers.
Yes, as I understand it, to consider a tangent line, the limit $x \to p$ must be taken for an interior point of the domain.

#### FactChecker

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As long as E is a field with a metric, I think you can probably go quite far with a general derivative definition. But I have not thought about it very much.

#### fresh_42

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As long as E is a field with a metric, I think you can probably go quite far with a general derivative definition. But I have not thought about it very much.

#### lavinia

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As long as E is a field with a metric, I think you can probably go quite far with a general derivative definition. But I have not thought about it very much.
You may like the Radon-Nikodym Theorem - a theorem in measure theory - which generalizes derivatives to derivatives of measures.

If $μ$ is a measure and $f$ is a positive measurable function then one gets a new measure $M(V)= ∫_{V}fdu$. The Radon Nikodym Theorem gives conditions under which one measure is an integral of a measurable function with respect to another measure.

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#### WWGD

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In Rudin, the derivative of a function $f: [a,b] \to \mathbb{R}$ is defined as:

Let $f$ be defined (and real-valued) on $[a,b]$. For any $x \in [a,b]$, form the quotient $\phi(t) = \frac{f(t) - f(x)}{t-x}\quad (a < t <b, t \neq x)$ and define $f'(x) = \lim_{t \to x} \phi(t)$, if the limit exists.

An analoguous definition is given for functions defined on an open interval.

Q1: Why do we restrict ourselves to defining derivatives on intervals? I would propose the following definition:

Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function and $x$ a limit point of $E$.

Define $$\phi: E \setminus \{x\} \to \mathbb{R}: t \mapsto \frac{f(t) - f(x)}{t-x}$$. Then we define $$f'(x) = \lim_{t \to x} \phi(t)$$ if the limit exists.

Q2: The quotient rule in Rudin is written in the following way:

Suppose $f,g$ are real functions on $[a,b]$ that are differentiable at $x \in [a,b]$. Then $f/g$ is differentiable at $x$, provided that $g(x) \neq 0$ and

$\left(\frac{f}{g}\right)'(x) = \frac{g(x)f'(x) - g'(x)f(x)}{g^2(x)}$

Following Rudin's convention, the domain of $f/g$ are those point $x$ of $[a,b]$ where $g(x) \neq 0$.
But if f/g is defined in the set where $g(x) \neq 0$ and given g(x) is continuous ( wherever it is differentiable) , then

EDIT
$[a,b]- g^{-1}( {0})$ is the set of differentiability. For most functions, I would think this is $[a,b]$ minus a discrete set. You do have functions where this is not true ( like, e.g., Test functions -- Smooth functions with compact support -- but I think at this point in Royden you will not likely get into these.

Ultimately, I think the set of points of differentiability is known to be, I think, an $F_{\sigma \delta}$ set. Shouldn't be too hard to check this out : https://math.stackexchange.com/questions/105810/continuous-functions-are-differentiable-on-a-measurable-set

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