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.
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!
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!
There is really a promising way to explain this and such technique is commonly used in analysis. The basic principle is that you want to express thus stuff in terms of the ones given, i.e. u’(x) and v’(x). To be more precise, [u(x+h)-u(x)]/h and the same thing for the function v. So in order to “inject” those given terms in u(x+h)v(x+h)-u(x)v(x) you might say well let’s do the first term and come up with [u(x+h)-u(x)]v(x+h)-u(x)v(x) however it’s no quite true for them to be equal such insertion pays cost and is not free you need to add u(x)v(x+h) back so that we don’t change the amount at all.
The people who proved it the first time, they did it through trial and error. It's about finding the right term that gives you what you want with out actually changing anything. Its somewhat similar to how you add, subtract or multiply a number on both sides to get rid of it on one side. The whole purpose of that is just so you can change the form of the equation that allows to to proceed to a meaningful result/solution .
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.
Experience with solving various mathematical problems helps one to see this as a possible need in a problem. The numerator had the look of a simplified algebraic expression, a clue to try to reverse that process. You are correct that the terms could have been reversed, what are the consequences of this decision? This is the kind of thinking one needs to have as a mathematician. Can you still find the result we are looking for? Cheerful Calculations! 🧮
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?
There is really a promising way to explain this and such technique is commonly used in analysis. The basic principle is that you want to express thus stuff in terms of the ones given, i.e. u’(x) and v’(x). To be more precise, [u(x+h)-u(x)]/h and the same thing for the function v. So in order to “inject” those given terms in u(x+h)v(x+h)-u(x)v(x) you might say well let’s do the first term and come up with [u(x+h)-u(x)]v(x+h)-u(x)v(x) however it’s no quite true for them to be equal such insertion pays cost and is not free you need to add u(x)v(x+h) back so that we don’t change the amount at all.
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. Reply
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.
Lots of good details and reasons given in the lecture.
DUDE YOU'RE A LIVE SAVER! THANK YOU!
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'
that was beautiful, I wish my teachers proved awesome shit like that
Love your channel eddie
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!
I just love you legend
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!
How does that "extra term" just pop out of thin air?
Like he said, it's only there so we can factor things out. That term appears through trial and error and experience
There is really a promising way to explain this and such technique is commonly used in analysis.
The basic principle is that you want to express thus stuff in terms of the ones given, i.e. u’(x) and v’(x). To be more precise, [u(x+h)-u(x)]/h and the same thing for the function v.
So in order to “inject” those given terms in u(x+h)v(x+h)-u(x)v(x) you might say well let’s do the first term and come up with [u(x+h)-u(x)]v(x+h)-u(x)v(x) however it’s no quite true for them to be equal such insertion pays cost and is not free you need to add u(x)v(x+h) back so that we don’t change the amount at all.
The people who proved it the first time, they did it through trial and error. It's about finding the right term that gives you what you want with out actually changing anything. Its somewhat similar to how you add, subtract or multiply a number on both sides to get rid of it on one side. The whole purpose of that is just so you can change the form of the equation that allows to to proceed to a meaningful result/solution .
You should also not that he is both adding and subtracting the same term which is basically 0 , any term + 0 -> term itself.
uv => vu'+uv' => OOF!
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.
Experience with solving various mathematical problems helps one to see this as a possible need in a problem. The numerator had the look of a simplified algebraic expression, a clue to try to reverse that process. You are correct that the terms could have been reversed, what are the consequences of this decision? This is the kind of thinking one needs to have as a mathematician. Can you still find the result we are looking for? Cheerful Calculations! 🧮
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?
There is really a promising way to explain this and such technique is commonly used in analysis.
The basic principle is that you want to express thus stuff in terms of the ones given, i.e. u’(x) and v’(x). To be more precise, [u(x+h)-u(x)]/h and the same thing for the function v.
So in order to “inject” those given terms in u(x+h)v(x+h)-u(x)v(x) you might say well let’s do the first term and come up with [u(x+h)-u(x)]v(x+h)-u(x)v(x) however it’s no quite true for them to be equal such insertion pays cost and is not free you need to add u(x)v(x+h) back so that we don’t change the amount at all.
It is not random, the reason is to help building the expression v(x+h)-v(x) / h
Hey nobody laughed on 'reproduce'...
Because they're not immature
@@tkay1937 also my comment was a sarcasm
Fuck
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.
Reply