my homework..

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- Sep 9th 2006, 07:04 AMshirelderivatives
my homework..

- Sep 9th 2006, 08:02 AMJones
There are rules/formulae for these these kinds of operations.

The addition formula, the product rule etc...

Just use them. - Sep 9th 2006, 01:26 PMCaptainBlack
- Sep 9th 2006, 01:34 PMCaptainBlack
Consider:

$\displaystyle

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

$

$\displaystyle

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

$

So:

$\displaystyle

\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:

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

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:) - Sep 9th 2006, 02:28 PMSoroban
Hello, shirel!

#3 asks us to derive the Product Rule.

. . Too bad that Latex is off-line right now.

Quote:

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

Calculate the derivative of $\displaystyle p(x)$.

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

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

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

$\displaystyle 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) $

$\displaystyle 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 $\displaystyle h:$

[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:

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

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

- Sep 9th 2006, 04:19 PMThePerfectHacker
There is just one problem Soroban.

Unlike, Captain**Blank**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). - Sep 10th 2006, 04:46 AMSoroban
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 . . .

- Sep 10th 2006, 06:25 AMThePerfectHacker
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).

- Sep 10th 2006, 09:25 AMPlato
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. - Sep 10th 2006, 09:30 AMThePerfectHacker
Come where?

Quote:

Originally Posted by**Plato**

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.

Quote:

Originally Posted by**Plato**

- Sep 10th 2006, 10:19 AMtopsquark
- Sep 10th 2006, 10:27 AMPlato
- Sep 10th 2006, 10:32 AMtopsquark
- Sep 10th 2006, 10:45 AMThePerfectHacker
- Sep 10th 2006, 04:12 PMPlato
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.