very well done. You explain this concept in simple and easy details. Far more understanding than most of my math professors. I appreciate the upload :)
The reason he can do that is because they cancel out and don't actually change the equation, they just make it easier to come to a solution. He uses those 2 expressions in order to factor out common terms and achieve the final goal but as you can see @2:42 is that the blue expressions just add up to 0, therefore leaving the expression undisturbed while being able to simplify it.
The limit as the number of rectangles in the summation approaches infinity IS the integral between two bounds; the antiderivative is an indefinite integral (no bounds).
By the way. . .Studying Electrical Engineering in a prestigious university in Canada. Just got done 2nd year and im done studying any new math material for the rest of my life (I think). So I wanted to personally thank Patrick. Probably wouldn'tve made it through Calc I.. .II. . .III. . Differential Equations. . .Linear Algebra and Vector Calculus without his help
Nice proof, I looked over one other video and was confused but this one explained it clearly enough to easily understand it. Have a good day if you’re reading this comment on a 10 y/o video OP
My name is Sergio Sanchez and i am a Mormon lol... i been watching your videos all day in preparation for an exam and this advertisement has been stuck in my head.
Actually wikipedia uses proper notation for it, also to understand the one on wikipedia you must also know that the difference quotient is the derivative with the addition of the limit as h approach 0.
Yes: the Riemann sum definition of the integral. You are essentially adding infinitesimal areas between the graph and the axis across the given interval a to b. All of calculus comes down to the concept of infinitesimal parts -- from this, and some clever math, everything can be derived.
Hey Patrick, I have a serious question. Have you heard of the Khan Academy? I think you and Salman could do so much good in the world since you're both really great teachers. I was thinking maybe you two could be partners and make videos that followed the typical curriculum for advanced math classes like at least college algebra to Calculus 3.
Yes I understand that, but as I've said before, the definite integral is the anti-derivative evaluated at the upper limit subtracted by the anti-derivative evaluated at the lower limit. I am aware of all of that, but I was asking about the proof of this or the process itself from which the definite integral was derived from the infinite Riemann summation.
Because it works. It's no different in principle to rationalizing a denominator or getting a common denominator when adding fractions. Adding something then subtracting it allows you to prove the result as well as maintain equality. It doesn't change the expression in regards to what it equals, only what it looks like.
TechTutorialsz If you read it from the start to the finish, it looks like it appears from nowhere, but if you read it from the end to the start, it just cancels out. It's basically the same thing but if you prefer one way or the other, the read it the way that helps you
Do you have guess of why randomly using f(x+h)g(x) works? It because of trial and error just because they took the product rule and work backwards to find the prove? Like the same thing for proving the integral of sec(x) how did the person know how to use sec(x)+tan(x) / sec(x) + tan (x)
Is there some sort of proof for why the definite integral of a function from a to b yields the area under its curve and is equal to its anti-derivative evaluated at B subtracted from it evaluated at A?
I know and understand the Riemann sum, but I don't understand is how they got from finding the limit of infinite rectangles to the concept of the antiderivative itself.
i have a really juvenile question. why is it that we can assume f(x+h) becomes f(x) when h->0 when we can also do so for basically everything else? Like g(x+h) and also the denominator h.. i understand that would make the whole equation untenable but why is it logically we can be selective about which term to bring to the limit, if you know what i mean
+Amanda C The limit of the product is the product of the limit. The limit as h approaches 0 is applied to f(x+h) because it can be factored out to be in its own limit. lim [f(x+h)*g(x+h)]=[lim f(x+h)]*[lim g(x+h)] (I excluded the denominator but you get the idea)
Thank you. I figured it was something like this but you may wan to explain that adding the expression f(x + h)g(x)... was just adding zero to your numerator since the expression will always equal zero and that doesn't change the value of the number in the numerator since it is zero.
This proof is good only if u suppose that derivative of f,g exist on "x" (i hope my maths english is not too bad lol) ,but ur explanation is very clear :)
Great presentation! I like the proof using Logs: Let y=f(x)g(x) and take the (natural) log of both sides ln(y)=ln(f(x)*g(x))=ln(f(x)) + ln(g(x)) ...Now differentiate both sides: (1/y)dy/dx=(1/f)df/dx +(1/g)dg/dx ... Solve for dy/dx ... dy/dx=y[(1/f)df/dx +(1/g)dg/dx] ...Now subs in for y ...i.e. y=f*g so dy/dx=f*g([(1/f)df/dx +(1/g)dg/dx] = g*df/dx + f/dg/dx QED
Not only that but unlike this proof it's not as cheaty. I don't like these proofs where it's like: "add this subtract this" because to come up with them you already need to know what you're looking for. The proof with differentials leads you to the product rule (and the chain rule) without requiring you to know it beforehand. As far as I know it's also the original proof by Leibniz which makes it even more interesting. :P
i dont understand the second to last step, if h never touches 0 then y does is equal 0? Because the second to last step equals 0 which is the prime definition. helppppppp
hey man do you still reply to comments? i just dont understand how to figure out the addition and subtraction of a term that's worth 0 you know where you added and subtracted *f(x+h)g(x)* at time frame 2:26 do you have exercises that goes deep into this and/or have explanation for techniques on this. i really want to understand this man
He picked that term because he knew how the end result looks like. If you don't know the end result then all you are trying to do is get factors of (f(x+h)-f(x))/h & (g(x+h)-g(x))/h since you know from there you can simplify it to f'(x) & g'(x).
just joking, i have heard of him. i am not interested in grants from people. then everything starts to become corporatized and you have to answer to some board of directors. no thanks, that is not for me.
This is excellent! A proof that can be presented in a high-school level class setting. Thank you very much. However, i do agree with one of the previous comments. The better and more intuitive one would be to use the partial differential chain rule which makes it pop right out. But this requires knowledge of partial differentiation which at least, my high school syllabus does not have. thank you.
Your proof is not complete. You have not proved that lim f(x+h) = f(x) as h goes to zero. This is true only because differentiability implies continuity. In general, it is false.
This was awesome! I love learning proofs after I've been using the rule, sort of justifies it. I appreciate you, Patrick.
@Micah Ryland reported,thank you
This is the first time a proof has really clicked for me and after following along I completely understood what is happening. Thank you for the video!
very well done. You explain this concept in simple and easy details. Far more understanding than most of my math professors. I appreciate the upload :)
The reason he can do that is because they cancel out and don't actually change the equation, they just make it easier to come to a solution. He uses those 2 expressions in order to factor out common terms and achieve the final goal but as you can see @2:42 is that the blue expressions just add up to 0, therefore leaving the expression undisturbed while being able to simplify it.
Man, I love this stuff. I struggled in calc in college, but watching these videos are very informative. Great channel!!
Honestly from the bottom of my heart, THANK YOU!
what was the thought process behind adding and subtracting f(x+h)g(x)
@that's fake but why they think to add and subtract it
The limit as the number of rectangles in the summation approaches infinity IS the integral between two bounds; the antiderivative is an indefinite integral (no bounds).
By the way. . .Studying Electrical Engineering in a prestigious university in Canada. Just got done 2nd year and im done studying any new math material for the rest of my life (I think). So I wanted to personally thank Patrick. Probably wouldn'tve made it through Calc I.. .II. . .III. . Differential Equations. . .Linear Algebra and Vector Calculus without his help
I started to have a nice grin the moment you added and subtracted f(x+h)*g(x). More proofs would be lovely!
Thank you for using 2 different colours it made it really eay to see where everything was going
Nice proof, I looked over one other video and was confused but this one explained it clearly enough to easily understand it. Have a good day if you’re reading this comment on a 10 y/o video OP
My name is Sergio Sanchez and i am a Mormon lol... i been watching your videos all day in preparation for an exam and this advertisement has been stuck in my head.
thanks man,we can never thank you enough but trust me you'll definitely be on my graduation speech :)
+Mlungisi Benzema10 IYASEBENZA LE CHAP BRO
+Pumlan Mamsundu Kakhulu futh mft
Wonderful Patrick. I didn't think I'd be able to follow it but you explained it beautifully. Many thanks.
Actually wikipedia uses proper notation for it, also to understand the one on wikipedia you must also know that the difference quotient is the derivative with the addition of the limit as h approach 0.
well, less important would be proofs to know. much more important are proof techniques: induction, proof by contradiction, etc
Yes: the Riemann sum definition of the integral. You are essentially adding infinitesimal areas between the graph and the axis across the given interval a to b. All of calculus comes down to the concept of infinitesimal parts -- from this, and some clever math, everything can be derived.
thanks for this i always forgot that sub in trick and never understood it great quick yet detailed video :D
Would be interesting to see you do a series on what you thought are the top ten most valuable proofs to learn in math education ? Top 20 ?
Hey Patrick, I have a serious question. Have you heard of the Khan Academy? I think you and Salman could do so much good in the world since you're both really great teachers. I was thinking maybe you two could be partners and make videos that followed the typical curriculum for advanced math classes like at least college algebra to Calculus 3.
Yes I understand that, but as I've said before, the definite integral is the anti-derivative evaluated at the upper limit subtracted by the anti-derivative evaluated at the lower limit. I am aware of all of that, but I was asking about the proof of this or the process itself from which the definite integral was derived from the infinite Riemann summation.
yes it is
Why do we substract and add "f(x+h)g(x)" Like is there a reason? or only we do it to make the proof work?
thats the only reason i came here and he just completely ignored an explanation.
Because it works. It's no different in principle to rationalizing a denominator or getting a common denominator when adding fractions. Adding something then subtracting it allows you to prove the result as well as maintain equality. It doesn't change the expression in regards to what it equals, only what it looks like.
TechTutorialsz If you read it from the start to the finish, it looks like it appears from nowhere, but if you read it from the end to the start, it just cancels out. It's basically the same thing but if you prefer one way or the other, the read it the way that helps you
Deuce1042 Thanks. I was wondering that, too.
TechTutorialsz watch this video it explains it clearly
And this is why I frickin' love maths.
It would be great if you could upload some theoretical stuff like this ! I mean, theorem proofs, etc Thaks Patrick !
You should try (imo) the differentials proof. It involves new concepts but it's much, much more intuitive.
I wish my teachers had gone over this in high school, instead of just teaching the short cut rules.
What gives you the "right" to factor out the terms f(x+h) and g(x) from the numerator h?
This is awesome, thank you so much. I wish you were my teacher...
i can run through them all if people want. it is pretty much the same idea in each!
Do you have guess of why randomly using f(x+h)g(x) works? It because of trial and error just because they took the product rule and work backwards to find the prove?
Like the same thing for proving the integral of sec(x) how did the person know how to use sec(x)+tan(x) / sec(x) + tan (x)
patrickjmt you're the math man!
Is there some sort of proof for why the definite integral of a function from a to b yields the area under its curve and is equal to its anti-derivative evaluated at B subtracted from it evaluated at A?
nice clear and concise video!
Awesome! Any chance of more advanced proofs, like proving that sup A is the lub of A?
this is superb sir...
I know and understand the Riemann sum, but I don't understand is how they got from finding the limit of infinite rectangles to the concept of the antiderivative itself.
Why did you add and substract that transaction, sir ?
i have a really juvenile question. why is it that we can assume f(x+h) becomes f(x) when h->0 when we can also do so for basically everything else? Like g(x+h) and also the denominator h.. i understand that would make the whole equation untenable but why is it logically we can be selective about which term to bring to the limit, if you know what i mean
+Amanda C The limit of the product is the product of the limit. The limit as h approaches 0 is applied to f(x+h) because it can be factored out to be in its own limit. lim [f(x+h)*g(x+h)]=[lim f(x+h)]*[lim g(x+h)] (I excluded the denominator but you get the idea)
f(x + h) approaches f(x) as h approaches 0. However, the limit of (f(x+h) - f(x))/h does not necessarily approach f(x).
the most primitive tools in the hands of a nobleman shine the light of wisdom and generosity.
Tengkiu patrick for making us understand easier
Thank you. I figured it was something like this but you may wan to explain that adding the expression f(x + h)g(x)... was just adding zero to your numerator since the expression will always equal zero and that doesn't change the value of the number in the numerator since it is zero.
This proof is good only if u suppose that derivative of f,g exist on "x"
(i hope my maths english is not too bad lol) ,but ur explanation is very clear :)
Great video. Really helped me
if they want to license / use my stuff, that is fine. but i do not think they have the least amount of desire or interest.
that was sooo cool lol
ty for this, wikipedia is so much harder to understand imo
Great presentation! I like the proof using Logs: Let y=f(x)g(x) and take the (natural) log of both sides ln(y)=ln(f(x)*g(x))=ln(f(x)) + ln(g(x)) ...Now differentiate both sides: (1/y)dy/dx=(1/f)df/dx +(1/g)dg/dx ... Solve for dy/dx ... dy/dx=y[(1/f)df/dx +(1/g)dg/dx] ...Now subs in for y ...i.e. y=f*g so dy/dx=f*g([(1/f)df/dx +(1/g)dg/dx] = g*df/dx + f/dg/dx QED
Why did you subtract and add the term? I mean. Which primciple states you can do that?
already have
why was the f(x+h)g(x) added then subtracted from the equation?
Very informative, thanks.
Not only that but unlike this proof it's not as cheaty. I don't like these proofs where it's like: "add this subtract this" because to come up with them you already need to know what you're looking for. The proof with differentials leads you to the product rule (and the chain rule) without requiring you to know it beforehand. As far as I know it's also the original proof by Leibniz which makes it even more interesting. :P
why do you move the limit sign to the right of the f(x+h) at 4:44 ? this makes no sense to me
because he plugged in the 0 into the h and got f(x)
The limit of a product is the same as the product of the limit of each term. So he was able to take the limit as h goes to 0 of f(x+h) which is f(x).
Distributing the limit between the two functions
Perfect!
Thanks.
you don't think that some are of particular fundamental value which serve as exemplars of the techniques?
Thank You Sir I`m from South Africa
can you run through proofs of all the calculus rules?
i dont understand the second to last step, if h never touches 0 then y does is equal 0? Because the second to last step equals 0 which is the prime definition. helppppppp
Can you do a video on Laplace transforms in the near future will appreciate it ..
why does f prime of x doesn't include dx any more? Does anybody know? Same question for g prime of x. Why doesn't it multiply with dx anymore?
never heard of him
Is the proof the same with that in a complex number sense?
tell me why? i dont get it. Like why are you adding and subtracting the same thing..is this trick...is it allowed...
basically how does the last step lead to the proof?
Patrick ! Thanks !!!
hey man
do you still reply to comments?
i just dont understand how to figure out the addition and subtraction of a term that's worth 0
you know where you added and subtracted *f(x+h)g(x)* at time frame 2:26
do you have exercises that goes deep into this and/or have explanation for techniques on this.
i really want to understand this man
He picked that term because he knew how the end result looks like. If you don't know the end result then all you are trying to do is get factors of (f(x+h)-f(x))/h & (g(x+h)-g(x))/h since you know from there you can simplify it to f'(x) & g'(x).
can someone please help I didn't understand his explanation when he explains on the limit f(x+h)-f(x) over h.
that's the definition of a derivative.
thanx dear.. am now enjoying everything is enlightened.
Why are you able to just add and subtract f(x+h)g(x)?
it doest change the function its like +2-2 its anyway equel to zero :)
ok i get that it doesnt change the function, but i would like to she how it was done.
thanku sooo much sir....now my problen is solve thanku.....😊
I have a question why you applied the limit to both sides of the sum. This near 4min
because you split it in two. it's a bit like:
x(a + b) = xa + xb
life saver, thanksssss
Excellent! Thanks.
you always safe my life
Thank you!
just joking, i have heard of him. i am not interested in grants from people. then everything starts to become corporatized and you have to answer to some board of directors. no thanks, that is not for me.
5+1-1+1-1+1-1=5 I can add and subtract 1 as many times as I'd like, I'm still not changing the equation.
i just did
just blithely made f(x+h) = f(x) on the left but not elsewhere.
thanks you sir
Thanks
This is excellent! A proof that can be presented in a high-school level class setting. Thank you very much.
However, i do agree with one of the previous comments. The better and more intuitive one would be to use the partial differential chain rule which makes it pop right out. But this requires knowledge of partial differentiation which at least, my high school syllabus does not have.
thank you.
Frighteningcar Thanks for the recommendation.
Good video
❤
what are the chances i just went over this today in class D:
nice. thank you
Perfect
Very good method.Thanks a lot.DrRahul Rohtak.India
It would NEVER cross my mind to add and subtract that :/
beautiful.
*me* : that makes sense
_*Also Me*_: (0.0) naw.
it is not at all insightful to be honest (like many good proofs), however it is correct.
@@patrickjmt *However* It is correct. End of the Statement. Lmao
Thank yuo
Your proof is not complete. You have not proved that lim f(x+h) = f(x) as h goes to zero. This is true only because differentiability implies continuity. In general, it is false.
f(x) is implied to be differentiable, therefore it's continuous, and you can just evaluate the limit by plugging in h=0.
ohhhhhhhhh yeah it makes sense
Wish I could switch brain with you...
💗💗💗💗💗💗
*ooooooh
woooooooooooooow
that's easy
Thank you!