Question

Calc I: lim(fg) = lim f lim g
but (fg) ' not = f ' g '

Shouldn't the above be equivalent since f ' = lim f ?

Answer 1

lim f has no meaning unless you specify what the limit is approaching

Answer 2

but either way lim f is not the same as f'

Answer 3

so what are yo asking for, the proof of the product rule?

Answer 4

you must remember that\[f'(x)=\lim_{h\to0}{f(x+h)-f(x)\over h}\]that is not just any old limit, but the derivative...

Answer 5

My teaches told us that 'derivative' is just another name for limit. Sum and constant multiple rules seem to match, but then product and power rules don't.... Still a bit confused, but tnx for help

Answer 6

your teacher sucks then, sorry

Answer 7

I will prove the product rule for you if you have the patience, but @Algebraic! may be doing it as we speak

Answer 8

...or not

Answer 9

I think we're assuming we're talking about the limit as the definition of the derivative.

Answer 10

right, but surely you agree that the limit and derivative are not names for the same thing; that is terrible terminology at best

Answer 11

the limit of a function as x approaches a is\[\lim_{x\to a}f(x)\]the derivative of a function at x=a is\[f'(a)=\lim_{h\to a}{f(x)-f(a)\over x-a}\]very different things....
you want the proof or not?

Answer 12

if you dont mind
or maybe a link will do
TNX

Answer 13

nah I'll do it :)

Answer 14

hugs and kisses...

Answer 15

The general formula for the derivative of a function by definition is\[\frac d{dx}f(x)=\lim_{h\to0}{f(x+h)-f(x)\over h}\]Now we want to know\[\frac d{dx}[f(x)g(x)]\]so we apply the same definition...\[\frac d{dx}[f(x)g(x)]=\lim_{h\to0}{f(x+h)g(x+h)-f(x)g(x)\over h}\]Now a trick...\[=\lim_{h\to0}{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)\over h}\]\[=\lim_{h\to0}{[f(x+h)-f(x)]g(x+h)+f(x)[g(x+h)-g(x)]\over h}\]\[=\lim_{h\to0}{f(x+h)-f(x)\over h}g(x+h)+f(x)\lim_{x\to0}{g(x+h)-g(x)\over h}\]\[\frac d{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)~~~~~~~~~~~~~~~~~~~~\huge\checkmark \]

Answer 16

I'm a perfectionist, so I cleaned it up
no typose though, so there you have it :)

Answer 17

typos* lol

Answer 18

TNX again

Answer 19

welcome!