Understanding the Derivative: A Beginner's Guide

[ILVB-loved]

someone said it was 20 but could not tell me how they arrived at that conclusion...

I know next to nothing about calculus but I am hoping by reading on this forum to have my brain primed to learn something of it in the days to follow.

Thank you for helping me out here (I am not a student, btw--just a seeker :-) )
ILVB

PS if this question falls under another thread subject that is alive and in progress would the administrator please determine the proper location for it and move it there? I don't want to start littering the forum by being a random thread generator/owner...thanks! :-)

 

[Mandelbroth]

someone said it was 20 but could not tell me how they arrived at that conclusion...

I know next to nothing about calculus but I am hoping by reading on this forum to have my brain primed to learn something of it in the days to follow.

Thank you for helping me out here (I am not a student, btw--just a seeker :-) )
ILVB

PS if this question falls under another thread subject that is alive and in progress would the administrator please determine the proper location for it and move it there? I don't want to start littering the forum by being a random thread generator/owner...thanks! :-)

No. The derivative of a constant is always 0.

 

[wifi]

More specifically, the derivative represents the rate of change of a function. This is why any constant function has a derivative of zero.

 

[ILVB-loved]

thank you. :-) I hope to learn the mechanics of calculus soon :-)

 

[berkeman]

thank you. :-) I hope to learn the mechanics of calculus soon :-)

You can start with this simple wikipedia introductory article:

http://en.wikipedia.org/wiki/Calculus

 

[Fredrik]

Nitpicking a bit: The derivative of the number 11 is undefined. The derivative of the function that takes every real number to 11, is the function that takes every real number to 0. This function is usually denoted by 0, but it's not the same thing as the number 0.

In other words, let f be the function defined by f(x)=11 for all x. Then f'(x)=0 for all x.

 

[Mandelbroth]

Nitpicking a bit: The derivative of the number 11 is undefined. The derivative of the function that takes every real number to 11, is the function that takes every real number to 0. This function is usually denoted by 0, but it's not the same thing as the number 0.

In other words, let f be the function defined by f(x)=11 for all x. Then f'(x)=0 for all x.

A constant is everywhere equal to its corresponding constant function. Thus, it is acceptable to say that every constant is, itself, a function that maps everything and the kitchen sink to that constant. That is,

$$C:\mathbb{E}\cup\left\{kitchen \, sink\right\}\rightarrow\{C\}$$ where $\mathbb{E}$ is the compliment of {kitchen sink} in the set of all possible inputs. Not all inputs to functions have to be real numbers, you know. They don't even have to be numbers at all.

I know what you're saying, and it IS correct, but I think saying that there is a difference between a constant and a function mapping any possible input to a constant is a little...ridiculous. It's easier to just treat every constant as a function with an output of its value.

 

[Fredrik]

Not all inputs to functions have to be real numbers, you know. They don't even have to be numbers at all.

Of course. But a person who asks about the derivative of 11 hasn't encountered any other kind of function.

I know what you're saying, and it IS correct, but I think saying that there is a difference between a constant and a function mapping any possible input to a constant is a little...ridiculous. It's easier to just treat every constant as a function with an output of its value.

I disagree. I think it's important to use accurate terminology from the start.

 

[jbunniii]

A constant is everywhere equal to its corresponding constant function. Thus, it is acceptable to say that every constant is, itself, a function that maps everything and the kitchen sink to that constant. That is,

$$C:\mathbb{E}\cup\left\{kitchen \, sink\right\}\rightarrow\{C\}$$ where $\mathbb{E}$ is the compliment of {kitchen sink} in the set of all possible inputs. Not all inputs to functions have to be real numbers, you know. They don't even have to be numbers at all.
.

Fair enough, but it doesn't make sense to talk about differentiability if the domain is just some abstract set. The domain must have enough structure (e.g., a normed vector space) so that the statement that "$f$ can be locally approximated by a linear map" makes sense. Additionally, in order to take a meaningful limit, it only makes sense to talk about differentiability at points of the domain which are limit points.

 

[Mandelbroth]

Fair enough, but it doesn't make sense to talk about differentiability if the domain is just some abstract set. The domain must have enough structure (e.g., a normed vector space) so that the statement that "$f$ can be locally approximated by a linear map" makes sense. Additionally, in order to take a meaningful limit, it only makes sense to talk about differentiability at points of the domain which are limit points.

The set of real numbers is a subset of the domain in question.

Since you seem to be interested, a normed vector space is not necessary. Consider differentiation on topological manifolds.

I disagree. I think it's important to use accurate terminology from the start.

Well, I suppose this has benefits. Then again, calculus sounds so much less complicated to people who are just starting to learn antidifferentiation if we say "plus a constant" instead of "plus any arbitrary element of the kernel of the derivative operator." I know I would have preferred the latter too, because the addition of a constant doesn't cover all possible antiderivatives, but sometimes people need time to adjust to new material.

This said, however, you have changed my mind on this. After some sleep, I've noticed that there are advantages to be had from more precise language, and a difference between a constant and a constant function allows for more precision in meaning.

 

[mathman]

someone said it was 20 but could not tell me how they arrived at that conclusion...

I know next to nothing about calculus but I am hoping by reading on this forum to have my brain primed to learn something of it in the days to follow.

Thank you for helping me out here (I am not a student, btw--just a seeker :-) )
ILVB

PS if this question falls under another thread subject that is alive and in progress would the administrator please determine the proper location for it and move it there? I don't want to start littering the forum by being a random thread generator/owner...thanks! :-)

It sounds like a garbled joke or riddle.

 

[micromass]

Consider differentiation on topological manifolds.

You can't take derivatives on topological manifolds.

 

[Mandelbroth]

You can't take derivatives on topological manifolds.

...wut?
A differentiable manifold is defined as a topological manifold with a globally defined differential structure.

 

[WannabeNewton]

But you said "differentiation on topological manifolds": the notion of differentiation in the category of topological manifolds is in and of itself not well defined because differentiation does not behave well under homeomorphisms alone. That is why we need a smooth structure; differentiation makes sense in a general context in the category of smooth/differentiable manifolds but we need that extra smooth structure; just having a topological manifold won't cut it. These are not abstract sets by any means as they have smooth structures; jbun simply gave normed vector spaces as an example (they are special cases of smooth manifolds anyways).

 

[Mandelbroth]

But you said "differentiation on topological manifolds": the notion of differentiation in the category of topological manifolds is in and of itself not well defined because differentiation does not behave well under homeomorphisms alone. That is why we need a smooth structure; differentiation makes sense in a general context in the category of smooth/differentiable manifolds but we need that extra smooth structure; just having a topological manifold won't cut it. These are not abstract sets by any means as they have smooth structures; jbun simply gave normed vector spaces as an example (they are special cases of smooth manifolds anyways).

I know this, and I am aware that smooth manifolds are not abstract sets (to my knowledge). It appeared to me that jbun is interested in math and would enjoy learning more of it. I was unsure of jbun's depth of knowledge in the topic, and I was only attempting to point in the general direction of manifolds (a topic that I find interesting), some of which are differentiable but not normed vector spaces.

I was not attempting to prove a point with this. I only wanted to encourage exploration. I'm pretty sure that's a goal for this site, right?

 

[micromass]

I know this, and I am aware that smooth manifolds are not abstract sets (to my knowledge). It appeared to me that jbun is interested in math and would enjoy learning more of it. I was unsure of jbun's depth of knowledge in the topic, and I was only attempting to point in the general direction of manifolds (a topic that I find interesting), some of which are differentiable but not normed vector spaces.

I was not attempting to prove a point with this. I only wanted to encourage exploration. I'm pretty sure that's a goal for this site, right?

In mathematics, it's crucial to be precise. This is something that distinguishes mathematicians from other branches: to be extremely precise. So if you say something about topological manifolds being differentiable, then you can expect that this will be corrected.

And rest assure, the depth of jbun's knowledge is very extensive!

 

[Mandelbroth]

In mathematics, it's crucial to be precise. This is something that distinguishes mathematicians from other branches: to be extremely precise. So if you say something about topological manifolds being differentiable, then you can expect that this will be corrected.

And rest assure, the depth of jbun's knowledge is very extensive!

Again, I was pointing toward a general topic. I'm aware that differentiation on topological manifolds doesn't make sense most of the time.

I've since figured this out about jbun.

 

[WannabeNewton]

Again, I was pointing toward a general topic.

Of course I personally knew what you were talking about from prior interaction but a general audience might not so it helps to be careful. It certainly is a very, very minor point though. It just might not be clear from context for everyone is all.