Proving Product Rule

[brianthibaudeau]

Great explanations. You do some great vids, which is keeping me fresh on my calculus. When I was in AP Calc here in the States, a classmate came up with a good mnemonic to help remember the product (and quotient) rule.

If you call your first function f(x) and your second function g(x), which is what I was taught, then your product rule becomes f' * g + g' * f. Or, as he said, "fig plus gif".

Then when it comes to the quotient rule, it becomes (f'g-g'f)/g^2, or "fig minus gif all over g squared). This is what helped me remember anyway.

[dennycyr]

DUDE YOU'RE A LIVE SAVER! THANK YOU!

[alejrandom]

that was beautiful, I wish my teachers proved awesome shit like that <3

[missghani]

I had to stop at 6:04 just to write that he has the quality of knowing ''whats gonna happen in few weeks or so'

[teodoirvanstaden]

I'm writing my first Calculus mid-year exam, my first year at university studying Physics and Mathematics. Guess what I'm watching? XD
Next up - Quotient rule and then Chain rule!

Thank you for the marvelous explanation Eddie Woo!

[TranquilSeaOfMath]

Lots of good details and reasons given in the lecture.

[anvithkakkera]

U fucking genius! Proofs r so cool, I wish I was smart enough to do Olympiad level proofs, but until then, proofs like u did will continue to fuckin blow my mind!

[IsmailPbx]

Who on earth thinks to add and subtract by the same thing. Why did you even put it in that order, you could have swapped around the minus part and the plus part.

[TheJProductins]

How does that "extra term" just pop out of thin air?

[faridthefadil]

why add and subtract u(x+h)v(x)? it seems so random i want to know why it is exactly in that form? And not any other? Does anyone have any sources I could read up?

[rishavjain]

Hey nobody laughed on 'reproduce'...

[qualquan]

Does the traditional thing zeroing "h" selectively. Classic cherry picking.
Easier if f(x+h) depicted as f(x)+df and g(x+h) as g(x)+dg
Then THEIR product is simply: f(x).g(x) +f(x).dg +g(x).df +df.dg
Then subtracting f(x).g(x) and dividing by dx gives:
f(x).dg/dx +g(x).df/dx +df.dg/dx
or f(x).g' + g(x).f' while the remaining term df.dg/dx is removed as it contains ALL the averaging error (not because it is zero)
BTW, since f(x+h) is f(x) + df
So f(x+h). g' = f(x).g' + df.g'
Then df.g' is eliminated as it contains averaging error and NOT because it is zero.



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