# Thread: Formula for Differentiation of Any Function

1. Many of us are familiar with the difference quotient, and know that with it alone we are able to derive much of differential calculus. However, the difference quotient doesn't work in the same way that many of the rules of differentiation do, being much more general and abstract in nature. Therefore, the point of this thread is to derive a formula less abstract, but as general as the conventional rules of differentiation for any function (using the difference quotient itself).

The first problem in such a procedure would be to analytically define such a function: for any given function of a real variable, I suggest that it can be defined as follows (if anyone has any objections please feel free to state so):

Therefore, by plugging this function into the difference quotient the result is as such:

Now, let us define as

To differentiate the function we can use the chain and product rule, namely the known formula .

Let us define as .

Therefore, are derivative amounts to .

QED

What the above essentially means is that any real function of a real variable can be written as , and

2.

3. Originally Posted by Ellatha
What the above essentially means is that any real function of a real variable can be written as , and
Of course it can, just take

Then and which should not be much of a surprise.

You have not really said anything.

4. Originally Posted by DrRocket
Of course it can, just take

Then and which should not be much of a surprise.

You have not really said anything.
That's only for functions of the form . I certainly do not think that there is an "of course" factor to it; g(x) can be any real valued function of a real variable, which includes , , , rather than the overly simple and convenient case of g(x) = 1.

EDIT: by the way, I've simplified down the function a bit, getting rid of Euler's number.

5. Originally Posted by Ellatha
Originally Posted by DrRocket
Of course it can, just take

Then and which should not be much of a surprise.

You have not really said anything.
That's only for functions of the form . I certainly do not think that there is an "of course" factor to it; g(x) can be any real valued function of a real variable,
No, it can't. won't work unless f is non-negative for instance. In any case you are chasing your tail in an unnecessarily complicated way.

Any function is of the form f(x). Have you been drinking ?

6. Originally Posted by DrRocket
No, it can't. won't work unless f is non-negative for instance.

Originally Posted by DrRocket
In any case you are chasing your tail in an unnecessarily complicated way.
I don't believe so; as you noted in the thread relating to proving the distributive property, there is a difference between providing a closed form of an expression and merely having a formula that allows you to find one (such as being able to use the chain rule in any case). This is similar to using the quadratic formula to solve for a quadratic as opposed to squaring the circle for any individual quadratic that comes along.

Originally Posted by DrRocket
Any function is of the form f(x). Have you been drinking ?
By that reasoning we should disregard the chain, product, and quotient rules since any function of the form , , is just a function . If you read the introductory paragraph, I mention that I'd like to find a formula for finding the derivative of a function without being overly abstract to the extent that it can't be used all that well, and thus is about as general as one can be.

I do agree though that with this formula one must still use the three aforementioned rules.

You can, for example, prove the power rule using the above formula:

,

The above proof is actually better than that using the binomial theorem originally developed by Isaac Newton, since it holds true for all n as an element of the real numbers, rather than for all n as an element for the natural numbers.

I believe that the current power rule should be replaced for rather than simply . Doing so would make it more generalized in addition to more compatible with the current product, quotient, and chain rules (which are expressed in terms of two functions f(x), g(x)). Although in an introductory calculus course it would probably still be more beneficial to teach the traditional x^n interpretation of the power rule (and than generalize it in advanced studies).

7. Originally Posted by Ellatha
Originally Posted by DrRocket
No, it can't. won't work unless f is non-negative for instance.

Which is WRONG unless f is non-negative.

Originally Posted by DrRocket
In any case you are chasing your tail in an unnecessarily complicated way.
Originally Posted by Ellatha
I don't believe so; as you noted in the thread relating to proving the distributive property, there is a difference between providing a closed form of an expression and merely having a formula that allows you to find one (such as being able to use the chain rule in any case). This is similar to using the quadratic formula to solve for a quadratic as opposed to squaring the circle for any individual quadratic that comes along.
This makes little sense.

Originally Posted by Ellatha
Originally Posted by DrRocket
Any function is of the form f(x). Have you been drinking ?
By that reasoning we should disregard the chain, product, and quotient rules since any function of the form , , is just a function . If you read the introductory paragraph, I mention that I'd like to find a formula for finding the derivative of a function without being overly abstract to the extent that it can't be used all that well, and thus is about as general as one can be.
ridiculous

Originally Posted by Ellatha
I do agree though that with this formula one must still use the three aforementioned rules.

You can, for example, prove the power rule using the above formula:

,

The above proof is actually better than that using the binomial theorem originally developed by Isaac Newton, since it holds true for all n as an element of the real numbers, rather than for all n as an element for the natural numbers.

I believe that the current power rule should be replaced for rather than simply . Doing so would make it more generalized in addition to more compatible with the current product, quotient, and chain rules (which are expressed in terms of two functions f(x), g(x)). Although in an introductory calculus course it would probably still be more beneficial to teach the traditional x^n interpretation of the power rule (and than generalize it in advanced studies).
There is no good vreason to do this.

You are indeed chasing your tail and making something that is quite simple look complicated. That is the opposite of good mathematics.

8. Originally Posted by DrRocket
Which is WRONG unless f is non-negative.
It is not "wrong," it just lead to a complex solution for all f(c) < 0. This is not a real-valued function. You might as well say that the power rule is wrong for n = 1/2 for x negative.

Originally Posted by DrRocket
This makes little sense.
It means that the rules of differentiation should be as such:

product rule: f(x)g(x)
chain rule: f(g(x))
quotient rule: f(x)/g(x)
power rule: f(x)^g(x)

because those rules are more general as opposed to those used today (which are the same except for the power rule being for x^n).

Originally Posted by Ellatha
ridiculous
Why? For any function f(x) one can use the difference quotient, yes, but by that method one would have to find the derivative rather than using the formula provided (which is a matter of plugging in).

Originally Posted by DrRocket
There is no good vreason to do this.

Originally Posted by DrRocket
You are indeed chasing your tail and making something that is quite simple look complicated. That is the opposite of good mathematics.
The work required to get to the formula is sophisticated, but the results (which are what are important), are rather simple, all they state is the following:

If anything, the current power rule may still be called the power rule, but the f(x)^g(x) method that I mention should be considered a generalized (and overall better) version of it. This is because it makes functions that would be extremely tedious to differentiate much less so (as they would require the chain rules as well as a variety of others). On the other hand, I will concede that slightly more simple functions, such as the case of x^n, require a little more work, but by that token one can simply derive a formula for such simple cases (as I already have shown by proving the current power rule).

9. Originally Posted by Ellatha
Originally Posted by DrRocket
Which is WRONG unless f is non-negative.
It is not "wrong," it just lead to a complex solution for all f(c) < 0. This is not a real-valued function. You might as well say that the power rule is wrong for n = 1/2 for x negative.
Which it is.

The square root of a negative number is not even well-defined until you have chosen a branch of the complex logarithm.

Originally Posted by Ellatha
Originally Posted by DrRocket
]This makes little sense.
It means that the rules of differentiation should be as such:

product rule: f(x)g(x)
chain rule: f(g(x))
quotient rule: f(x)/g(x)
power rule: f(x)^g(x)

because those rules are more general as opposed to those used today (which are the same except for the power rule being for x^n).
silly

Unncessarily complicated and do not carry over to the comples case unambiguously until you pick a branch of the logarithm.

Originally Posted by Ellatha
Originally Posted by DrRocket
ridiculous
Why? For any function f(x) one can use the difference quotient, yes, but by that method one would have to find the derivative rather than using the formula provided (which is a matter of plugging in).

Originally Posted by DrRocket
There is no good vreason to do this.

Originally Posted by DrRocket
You are indeed chasing your tail and making something that is quite simple look complicated. That is the opposite of good mathematics.
The work required to get to the formula is sophisticated, but the results (which are what are important), are rather simple, all they state is the following:

If anything, the current power rule may still be called the power rule, but the f(x)^g(x) method that I mention should be considered a generalized (and overall better) version of it. This is because it makes functions that would be extremely tedious to differentiate much less so (as they would require the chain rules as well as a variety of others). On the other hand, I will concede that slightly more simple functions, such as the case of x^n, require a little more work, but by that token one can simply derive a formula for such simple cases (as I already have shown by proving the current power rule).
This entire discussion is silly.

You started out with in which case you may as well take , which will be necessary if f has a log.

On the other hand in general is DEFINED to be when the necessary expressions are defined, and the usual methods of differentiation are quite adequate in that case.. There is no need to memorize anything and no "rules" are needed.

10. Originally Posted by DrRocket
Which it is.

The square root of a negative number is not even well-defined until you have chosen a branch of the complex logarithm.
Okay; it's just a matter of your solution not being in the designated set of numbers rather than a problem with the formula than; I don't view this as an issue.

Originally Posted by DrRocket
silly

Unncessarily complicated and do not carry over to the comples case unambiguously until you pick a branch of the logarithm.
Apparently mathematicians at leading universities have chosen to teach three of the four rules provided to their students, so they obviously are useful and warrant consideration (and I can only imagine you use them yourself when differentiating a function). There is no flaw in the f(x)^g(x) formula that can't be found in x^n, therefore all of your arguments relating to the set of complex numbers being a possible solution do not suggest that the current system in use is better than that for which I have proposed.

Originally Posted by DrRocket
This entire discussion is silly.

You started out with in which case you may as well take , which will be necessary if f has a log.
I assume you mean . A logarithm would be defined as such:

Why would g(x) need to equal one? Furthermore, the rule allows one to differentiate any function for g(x) which makes it obviously useful to mathematics. You are purposely nitpicking various theoretical definitions while ignoring obvious applications for no particular reason.

Explain to us, please, how you would differentiate a function of the following form (in terms of an algorithm):

Originally Posted by DrRocket
On the other hand in general is DEFINED to be when the necessary expressions are defined, and the usual methods of differentiation are quite adequate in that case.. There is no need to memorize anything and no "rules" are needed.
Well the rules are practical and what are used by mathematicians to differentiate functions, they obviously have meaning and are not the result of tautological statements with which professors use to humor their students and purposely make them unprepared for real-world mathematics as you are ridiculously suggesting.

11. I'm tired of arguing this with you. Some of what you have said is wrong. Some of it is obvious. Most of the remainder is inane.

12. Originally Posted by DrRocket
I'm tired of arguing this with you. Some of what you have said is wrong. Some of it is obvious. Most of the remainder is inane.
1. is not obvious.

2. It is useful for differentiating a large group of functions.

Your main, predetermined point seems to be to suggest that I can't come up with anything satisfying the conditions of being useful and original at this stage in growth. From what I've read about Descartes he had similar preconceptions regarding the young Pascal (and his issue with Pascal being non-personal, therefore suggesting any young intellectual). Descartes is often regarded as blinding his intellect with his emotions. You should give young minds a chance: throughout history they have been an aid to professionals (consider Gauss, Pascal, Leibniz, Grotius, Laplace, von Goethe, Mozart, and many others).

I'm not saying I proved the Riemann Hypothesis, only that I've discovered a more generalized formula for differentiation than the preexisting one and that it may be of an aid for proving simpler ones and differentiating more sophisticated ones.

13. Originally Posted by Ellatha
Originally Posted by DrRocket
I'm tired of arguing this with you. Some of what you have said is wrong. Some of it is obvious. Most of the remainder is inane.
1. is not obvious.

2. It is useful for differentiating a large group of functions.

Your main, predetermined point seems to be to suggest that I can't come up with anything satisfying the conditions of being useful and original at this stage in growth. From what I've read about Descartes he had similar preconceptions regarding the young Pascal (and his issue with Pascal being non-personal, therefore suggesting any young intellectual). Descartes is often regarded as blinding his intellect with his emotions. You should give young minds a chance: throughout history they have been an aid to professionals (consider Gauss, Pascal, Leibniz, Grotius, Laplace, von Goethe, Mozart, and many others).

I'm not saying I proved the Riemann Hypothesis, only that I've discovered a more generalized formula for differentiation than the preexisting one and that it may be of an aid for proving simpler ones and differentiating more sophisticated ones.

so it is quite obvious from the more usual expression. One simple algebra step of dubious value. This qualifies as obvious, inane or perhaps both.

Remember that it is meaningful only when f is non-negative, unless one has chosen a branch of the logarithm. That important fact is somewhat obscured by your "rule".

You have discovered nothing, and you would do well to quit patting yourself on the back and recognize that fact.

What you have done is find a rather long and clumsy way to get to a simple result. You seem to have spent too much time in an education system that is more concerned with your self-esteem than with substantive education.

14. Originally Posted by Ellatha
Descartes is often regarded as blinding his intellect with his emotions. You should give young minds a chance: throughout history they have been an aid to professionals (consider Gauss, Pascal, Leibniz, Grotius, Laplace, von Goethe, Mozart, and many others).
1. The body of mathematics was much smaller.
2. Pascal was a hell of a lot smarter.

15. Originally Posted by DrRocket

so it is quite obvious from the more usual expression. One simple algebra step of dubious value. This qualifies as obvious, inane or perhaps both.
You are purposely confusing obliviousness with simplicity. The three introductory lines in your above post are sufficient to describe most of what I stated in my original post, although I wanted to explain my points thoroughly and it's always in one's best interest to show all of their work. The result is always the most important aspect, however, and you have yet to provide a reason as to why it is flawed and unworthy of use.

Originally Posted by DrRocket
Remember that it is meaningful only when f is non-negative, unless one has chosen a branch of the logarithm. That important fact is somewhat obscured by your "rule".
That is if you consider meaning to be any derivative in the set of real numbers. This point does not justify some of the claims that you have made against the formula (as they can easily be applied to the ones currently in use).

Originally Posted by DrRocket
You have discovered nothing, and you would do well to quit patting yourself on the back and recognize that fact.

Originally Posted by DrRocket
What you have done is find a rather long and clumsy way to get to a simple result.
There exists no clumsiness as far as reasoning is concerned; a conclusion is either true or not depending on the reasons provided. The explanation for the duration of the work is the result of an emphasis to show one's full work (including simplifications).

Originally Posted by DrRocket
You seem to have spent too much time in an education system that is more concerned with your self-esteem than with substantive education.
I wouldn't know; I slept through most of my classes.

Originally Posted by DrRocket
1.The body of mathematics was much smaller.
Reasoning works the same today as it did in the time of the dinosaurs.

Originally Posted by DrRocket
2. Pascal was a hell of a lot smarter.
No.

16. Originally Posted by Ellatha
Originally Posted by DrRocket
2. Pascal was a hell of a lot smarter {than you}.
No.
I cannot believe you said that!! If you truly think so, it will be a job for the men in white coats.

17. Originally Posted by Guitarist
I cannot believe you said that!
It's in writing and I haven't edited my post, so believe it.

18. Originally Posted by Ellatha
Originally Posted by Guitarist
I cannot believe you said that!
It's in writing and I haven't edited my post, so believe it.
So are your attempts at mathematical proof.

You are deluding yourself.

That is not good. Failue to live up to expectations has been ruinous to mathematicians in the past, even real genuises. You are not one.

19. Originally Posted by DrRocket
...
"It is not possible on that basis to assess [']aptitude['], and assessing aptitude is hard even in person."

20. Originally Posted by Ellatha
Originally Posted by DrRocket
...
"It is not possible on that basis to assess [']aptitude['], and assessing aptitude is hard even in person."
No, but is quite easy to recognize foolishness that demonstrates lack of aptitude, particularly if one understands the subject matter well, has a lot of experience, has met real genuises and is familiar with the work of others..

Genuises that I have seen in person:

Rene Thom
Don Zagier
John Milnor
Shizuo Kakutani

Some other well-known mathematicians:

Lou Auslander
Ytzak Katznelson
George (Dan) Mostow
Cal Moore
Joe Wolf
Roger Howe
Jaque Tits
Harry Furstenberg
Hugo Rossi
Paul Sally
Paul Halmos
John Brezin

You are not even close. You might, with a lot of work, become a competent mathematician, but genuis would have been evident by now. It is not.

BTW even the genuises had to work very hard to produce the work for which they are known.

21. Your disgusting and tasteless attempts at changing the subject matter to me because your arguments failed is immature and embarrassing.

22. Originally Posted by Ellatha
Your disgusting and tasteless attempts at changing the subject matter to me because your arguments failed is immature and embarrassing.
You changed the subject.

None of my points are invalid, and the rationale behind them most certainly did not "fail".

Do you have any idea what you are talking about ?

If you are going to make progress you had better get your ego under control. You are not a genius, but most people have to live with that. There are no genuises in many very good mathematics departments.

23. Originally Posted by DrRocket
If you are going to make progress you had better get your ego under control.
Do not worry about my personal traits.

Originally Posted by DrRocket
Do you have any idea what you are talking about ?
I don't want your opinion on "whether I'm a genius or not." I never asked you, nor did I ask Guitarist. Quite frankly I found that comment an attempt at a bait such that any reply I generated would have caused an uproar from at least one member of this sub-forum. Considering many of the nitpicking arguments you posed in addition to your bellicose demeanor I believe it is not all that great of a violation.

Originally Posted by DrRocket
None of my points are invalid, and the rationale behind them most certainly did not "fail".
You provided no rationale for your points, as doing so would give you the impression that you were giving the impression that you actually read my post and considered what I stated.

All I really did was the following:

1. Demonstrated what the derivative of f(x)^g(x) was.
2. Suggested that the derivative was more useful than the current power rule, x^n in some aspects because it is more general (and actually allows one to prove the power rule).

Seemingly anyone that looked at those two points would not say that there is anything wrong with them and that they are true (even professional mathematicians). Not you, though; you had to go out of your way to find some sort of "flaw" that can be attributed to seemingly anything and make remarks about my competence for no reason whatsoever.

See the opening post in the following thread for more: http://www.thescienceforum.com/Hospi...eem-28561t.php

Originally Posted by DrRocket
You changed the subject.
Not to matters that related outside of this thread; my comments regarding you was the reason you even posted in the thread and resulted in the discussion. Your comments regarding genius are simply random.

24. [quote="Ellatha"]
I don't want your opinion .................quote]

We're done.

25. Originally Posted by DrRocket
We're done.
Good.

26. Originally Posted by Ellatha
Originally Posted by DrRocket
Which is WRONG unless f is non-negative.
It is not "wrong," it just lead to a complex solution for all f(c) < 0. This is not a real-valued function. You might as well say that the power rule is wrong for n = 1/2 for x negative.

Originally Posted by DrRocket
This makes little sense.
It means that the rules of differentiation should be as such:

product rule: f(x)g(x)
chain rule: f(g(x))
quotient rule: f(x)/g(x)
power rule: f(x)^g(x)

because those rules are more general as opposed to those used today (which are the same except for the power rule being for x^n).

Originally Posted by Ellatha
ridiculous
Why? For any function f(x) one can use the difference quotient, yes, but by that method one would have to find the derivative rather than using the formula provided (which is a matter of plugging in).

Originally Posted by DrRocket
There is no good vreason to do this.

Originally Posted by DrRocket
You are indeed chasing your tail and making something that is quite simple look complicated. That is the opposite of good mathematics.
The work required to get to the formula is sophisticated, but the results (which are what are important), are rather simple, all they state is the following:

If anything, the current power rule may still be called the power rule, but the f(x)^g(x) method that I mention should be considered a generalized (and overall better) version of it. This is because it makes functions that would be extremely tedious to differentiate much less so (as they would require the chain rules as well as a variety of others). On the other hand, I will concede that slightly more simple functions, such as the case of x^n, require a little more work, but by that token one can simply derive a formula for such simple cases (as I already have shown by proving the current power rule).
true

27. Originally Posted by Susanna Omori
true
False.

Better go back and read the whole thread, including the "derivation" of Ellatah's formula with a simple bit of algebra from the more usual approach.

There is no need for some complex formula, of limited utility and validity, when it follows simply from that which is well known.

28. Originally Posted by DrRocket
Originally Posted by Susanna Omori
true
False.

Better go back and read the whole thread, including the "derivation" of Ellatah's formula with a simple bit of algebra from the more usual approach.

There is no need for some complex formula, of limited utility and validity, when it follows simply from that which is well known.
You have no idea what you're talking about.

On a more general note, it would be really, REALLY nice if we all here (I include myself as an occasionally guilty party) could refrain from insulting comments, especially when there is no other content.

30. Originally Posted by Guitarist