Question

Could someone provide a step by step way to prove the product rule?

Answer 1

Why don't we <puts on shades>
@UseInduction ?

AWW YEAA

haha... couldn't resist, sorry. ;)
You still there?

Answer 2

lol yeah

Answer 3

Okay.... let f and g be differentiable functions... we're trying to find...
\[\LARGE \frac{d}{dx}f(x)g(x)\]
savvy? :D

Answer 4

yeah i know the rule itself and how to use it but the problem is the general proof of it and how to get there :)

Answer 5

So, by definition
\[\LARGE \frac{d}{dx}f(x)g(x)=\lim_{h \rightarrow0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\]I bet you got this far, at least ^.^

Answer 6

yeah, that's by following the general definition of derivative. got it so far

Answer 7

Well, how did you proceed from here?

Answer 8

are you supposed to just simplify the definition of derivative? :O

Answer 9

thanks if that is true, didn't think of it before you pointed it out :)

Answer 10

Unfortunately, not that simple :)
First, I need you to accept this... but it's almost trivial...
\[\LARGE \lim_{h \rightarrow 0}f(x+h)=f(x)\]

Answer 11

Any questions so far, @UseInduction ?

Answer 12

nope please continue

Answer 13

Now, we still have to prove that it's equal to
f'(x)g(x) + f(x)g'(x)
So... What to do...
Let's subtract AND add f(x+h)g(x) in the numerator, shall we?
\[\large \lim_{h \rightarrow0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\]

Answer 14

Yes? You have a comment and/or a query?

Answer 15

what is the reasoning behind the addition and subtraction?

Answer 16

You'll see. Now, we factor out some terms... by the way, I'll omit the limit part, because it's getting too bulky.
\[\Large \lim_{h \rightarrow 0}\large \frac{f(x+h)[g(x+h)-g(x)]+[f(x+h)-f(x)]g(x)}{h}\]
Following so far? Maybe you can now see why we added and subtracted that term ^.^

Answer 17

Huh... it seems I didn't omit the limit part after all... my bad :D

Answer 18

isnt it to simplify the division by h by getting all terms in the numerator with an h in them

Answer 19

Nope ;)

Answer 20

o well, keep going then :P

Answer 21

So, this time I'll omit the limit part, temporarily...
That fraction thingy can be split into two fractions...
\[\Large f(x+h)\cdot\frac{g(x+h)-g(x)}{h}+\frac{f(x+h)-f(x)}{h}\cdot g(x)\]Right?

Answer 22

yup

Answer 23

Can't you see it yet?
Limit distributes over addition, so...
\[\Large \lim_{h\rightarrow0}\left[ f(x+h)\cdot\frac{g(x+h)-g(x)}{h}+\frac{f(x+h)-f(x)}{h}\cdot g(x)\right]\]\[\Large \lim_{h\rightarrow0}\left[ f(x+h)\cdot\frac{g(x+h)-g(x)}{h}\right]+\lim_{h\rightarrow0}\left[\frac{f(x+h)-f(x)}{h}\cdot g(x)\right]\]

Answer 24

Sorry
\[\Large \lim_{h\rightarrow0}\left[ f(x+h)\cdot\frac{g(x+h)-g(x)}{h}+\frac{f(x+h)-f(x)}{h}\cdot g(x)\right]\]


\[\Large \lim_{h\rightarrow0}\left[ f(x+h)\cdot\frac{g(x+h)-g(x)}{h}\right]+\lim_{h\rightarrow0}\left[\frac{f(x+h)-f(x)}{h}\cdot g(x)\right]\]

Answer 25

ohh i see :) the post before the last post is f(x)g(x)' + f'(x)g(x)

Answer 26

And surely, you can see where this is going now? :)

Answer 27

Well, let me just patch things up...
\[\large \lim_{h\rightarrow0}f(x+h)\cdot\frac{g(x+h)-g(x)}{h}=\lim_{h \rightarrow 0}f(x+h)\cdot\lim_{h \rightarrow 0}\frac{g(x+h)-g(x)}{h}\]
Since both limits exist... and this is just equal to
\[\Huge f(x)\cdot g'(x)\]right, @UseInduction do something similar to the other term too ;)

Answer 28

\[\lim_{x \rightarrow 0} g(x+h) * \lim_{x \rightarrow 0}\frac{ f(x + h) - f(x)}{ h }\]

which is equal to g(x) * f '(x). Is this what you meant?

Answer 29

well, almost. Actually, it was
\[\large \frac{f(x+h)-f(x)}{h}\cdot g(x)\]
There was no g(x+h) involved :P

Answer 30

close enough