my homework..
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
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 $\displaystyle p(x) = f(x)\cdot g(x)$
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: .$\displaystyle p'(x)\;=\;f(x)\cdot g'(x) + g(x)\cdot f'(x)$
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).
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.
Come where?
Maybe, you taught an advanced class. Because I realized teachers do not like to get too theoretic about limits, they treat them more intuitevly.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.
I know, that is the joke.Originally Posted by Plato
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.