derivatives

**shirel** · September 9th 2006, 07:04 AM

my homework..

**Jones** · September 9th 2006, 08:02 AM

There are rules/formulae for these these kinds of operations.

The addition formula, the product rule etc...
Just use them.

**CaptainBlack** · September 9th 2006, 01:26 PM

Originally Posted by Jones

There are rules/formulae for these these kinds of operations.

The addition formula, the product rule etc...
Just use them.

He is being asked to prove them

RonL

**CaptainBlack** · September 9th 2006, 01:34 PM

Originally Posted by shirel

my homework..

1. consider the sum of two functions s(x): =f(x) + g(x) as a new function. calculate the derivative of s(x) and prove thereby that it equals the sum of the derivatives of g and g.

Consider:

$ \frac{s(x+h)-s(x)}{h}=\frac{f(x+h)+g(x+h)-f(x)-g(x)}{h} $

$ =\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h} $

So:

$ \lim_{h \to 0}\frac{s(x+h)-s(x)}{h}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} + \lim_{h \to 0}\frac{g(x+h)-g(x)}{h} $

which if all of these limits exist proves that:

$s'(x)=f'(x)+g'(x)$ .

The others are similar, just use the definition of what a derivative is.
Replace the function by its other representation and rearrange to the
form we know it must have.

RonL

The meaning of this will become clear (as will my TeX errors) when the
LaTeX system is reinstalled

**Soroban** · September 9th 2006, 02:28 PM

Hello, shirel!

#3 asks us to derive the Product Rule.
. . Too bad that Latex is off-line right now.

3. Consider the product of two functions $p(x) = f(x)\cdot g(x)$
Calculate the derivative of $p(x)$ .

We will derive: . $p'(x)\:=\:\lim_{h\to0}\frac{p(x+h) - p(x)}{h}$

We have: . $[(x+h) - p(x)\;=\;f(x+h)\cdot g(x+h) - f(x)\cdot g(x)$

Subtract and add $f(x+h)\cdot g(x):$

$p(x+h) - p(x)\;=\;f(x+h)\cdot g(x+h) - f(x+h)\cdot g(x) - f(x)\cdot g(x) + f(x+h)\cdot g(x)$

$p(x+h) - p(x)\;=\;f(x+h)\left[g(x+h) - g(x)\right] + g(x)\left[f(x+h) - f(x)\right]$

Divide by $h:$

[tex]\frac{p(x+h) - p(x)}{h} \;= \;f(x+h)\left[\frac{g(x+h)-g(x)}{h}\right] + g(x)\left[\frac{f(x+h)-f(x)}{h}\right] [/ math]

Take the limit:

$p'(x)\;=\;\lim_{h\to0}f(x+h)\cdot\lim_{h\to0}\left[\frac{g)x+h) - g(x)}{h}\right] + g(x)\cdot\lim_{h\to0}\left[\frac{f(x+h)-f(x)}{h}\right]$

Therefore: . $p'(x)\;=\;f(x)\cdot g'(x) + g(x)\cdot f'(x)$

**ThePerfectHacker** · September 9th 2006, 04:19 PM

There is just one problem Soroban.
Unlike, CaptainBlank who states that the limits exists and thus does their sum, you did not. In front of mathematicians your proof would not be acceptable. (This is the major difference between Calculus and Analysis).

**Soroban** · September 10th 2006, 04:46 AM

Hello, TPHacker!

Silly me . . . I thought this was a Calculus I problem.

I wasn't aware that this was a grad-school Advanced Analysis problem
. . where Existence must be established first.

I won't make that mistake again . . .

**ThePerfectHacker** · September 10th 2006, 06:25 AM

Originally Posted by Soroban

Hello, TPHacker!

Silly me . . . I thought this was a Calculus I problem.

I do not know if you are joking or not but there is no way this can be a Calculus I problem. The professor is not going to make you prove the product rule, which Leibniz was struggling to prove. I assume the students were familiar with the infromal proof from Calculus I but wanted them to show existence of limits. (Honestly how many people can spot that add-subtract trick).

**Plato** · September 10th 2006, 09:25 AM

Oh, come Perfect! It was proved in very Calculus I class I ever taught. It is a standard way of illustrating that the ‘derivative limit’ does not have to be written in the order we all are used to seeing.

BTW: It usually goes, “There are 10 kinds of people: those who understand binary numbers and those who don’t.

**ThePerfectHacker** · September 10th 2006, 09:30 AM

Originally Posted by Plato

Oh, come Perfect!

Come where?

Originally Posted by Plato

It was proved in very Calculus I class I ever taught.

Maybe, you taught an advanced class. Because I realized teachers do not like to get too theoretic about limits, they treat them more intuitevly.

I understand what you are saying that the proof is not so difficult, yes, I agree. I am saying that you need to show the two functions are differenciable and use rules about the existence of limits otherwise the proof is not sound.

Originally Posted by Plato

BTW: It usually goes, “There are 10 kinds of people: those who understand binary numbers and those who don’t.

I know, that is the joke.

**topsquark** · September 10th 2006, 10:19 AM

Originally Posted by ThePerfectHacker

Maybe, you taught an advanced class. Because I realized teachers do not like to get too theoretic about limits, they treat them more intuitevly.

I understand what you are saying that the proof is not so difficult, yes, I agree. I am saying that you need to show the two functions are differenciable and use rules about the existence of limits otherwise the proof is not sound.

As it happens, when I did my undergrad proofs like these were usually saved for the "Advanced Calculus" class in the fourth semester.

-Dan

**Plato** · September 10th 2006, 10:27 AM

Originally Posted by topsquark

As it happens, when I did my undergrad proofs like these were usually saved for the "Advanced Calculus" class in the fourth semester.

But in an advanced calculus class I would hope that you gave epsilon-delta proof for these properties. Not simple limit proofs. Now there I do agree: epsilon-delta proofs have no place in beginning calculus.

**topsquark** · September 10th 2006, 10:32 AM

Originally Posted by Plato

But in an advanced calculus class I would hope that you gave epsilon-delta proof for these properties. Not simple limit proofs. Now there I do agree: epsilon-delta proofs have no place in beginning calculus.

Actually, I never took the class so I can't say for sure. But I believe you are right about it.

-Dan

**ThePerfectHacker** · September 10th 2006, 10:45 AM

Originally Posted by Plato

But in an advanced calculus class I would hope that you gave epsilon-delta proof for these properties. Not simple limit proofs. Now there I do agree: epsilon-delta proofs have no place in beginning calculus.

Once you have extablished the basic rules of limits through delta-epsilon you can stop using it for more advanced limit proofs.

**Plato** · September 10th 2006, 04:12 PM

Originally Posted by ThePerfectHacker

Once you have extablished the basic rules of limits through delta-epsilon you can stop using it for more advanced limit proofs.

Not sure what you point is? But I can tell you that over a thirty year stint of teaching real analysis, undergraduate and graduate, we never used anything but the epsilon-delta proofs. Limits, continuity, derivatives, and integrals are all defined using epsilon-deltas. We insist, well at least the school of mathematics I represent, a solid proof using epsilon and deltas is the only correct proof.

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