Chain rule proof | Derivative rules | AP Calculus AB | Khan Academy

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Here we use the formal properties of continuity and differentiability to see why the chain rule is true.
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AP Calculus AB on Khan Academy: Bill Scott uses Khan Academy to teach AP Calculus at Phillips Academy in Andover, Massachusetts, and heÕs part of the teaching team that helped develop Khan AcademyÕs AP lessons. Phillips Academy was one of the first schools to teach AP nearly 60 years ago.
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[@dwong7826]

As several people pointed out, this proof is incorrect because it is incomplete. By dividing by delta u, delta u can't be zero. As viewers will note, the proof in the video proceeds assuming delta u is not zero -- an unstated assumption. The division by zero case is addressed in the Wikipedia article, en.wikipedia.org/wiki/Chain_rule.

[@stopthephilosophicalzombie9017]

Thank you for mentioning this. That problem was glaring out at me at the end.

[@illumexhisoka6181]

But it doesn't matter because if delta u is zero then the u function is a constant and we know without a proof that its Derivative is zero and when we put it inside another function it will also have a Derivative that = to zero

[@DarinBrownSJDCMath]

@@illumexhisoka6181 When we say "delta u is zero", we only mean that every interval of the point contains points in which delta u is zero, NOT that delta u is identically zero in some neighborhood of the point.

[@hamez1300]

@@DarinBrownSJDCMath I'm quite confused, is this in the case of rapidly oscillating functions? If so how do you resolve this?

[@DarinBrownSJDCMath]

@@hamez1300 The usual example given is u(x) = x^2 * sin(1/x) for x =/= 0, and u(0) = 0. Then delta(u) --> 0 as delta(x) --> 0, every interval about 0 contains points where delta(u) = 0, yet delta(u) is not identically 0 near 0. This causes the intuitive proof not to work, because delta(y)/delta(u) is undefined for points arbitrarily close to 0. There are several ways to "repair" this defect. The most common is to "redefine" delta(y)/delta(u) to be equal to y'(u(a)) whenever delta(u) = 0. I have always found this approach to be very ugly and not too instructive. My preferred repair is to use an equivalent definition of differentiability based on linear approximation error, which avoids division altogether. This alternate method has the additional advantage of generalizing nicely to the multivariable case.

[@gothicknight5538]

Thank you. I always love to know the proofs and not just simply memorizing it.

[@chempoint5854]

We r same bro

[@salmansaadiqmusic]

Me too man

[@user-pd4wz1oo3x]

not knowing the proof is kinda like not knowing the language the song you listen to is written in

[@Kneecap22]

The khan academy proof is not rigorous, I like the linear approximation and then turning dy into a function of the errors of W and then dW as a function of x by linear approximation and then substitute dW and the starting points, where the starting point of the inner function B is the outer function at A, and then divide dx, and then FOIL and then take the limit as dx goes to 0, so the error of linear_approximation_1 and linear_approximation_2, goes to 0. And you are left with (f'(g(a)))*(g'(a)).

[@12345papad]

This is not a proof. The proof of this requires either non-standard analysis ( which you seem to be assuming to hold true in this case) or the FTC.

[@seandever2272]

But since Δy = y(x+Δx)-y(x) isn't "the limit as Δx approaches zero of Δy over Δx" just a slightly simplified version of the FTC?

[@mariomario-ih6mn]

I know

[@jackmortem4557]

Hi

[@carlosmehicano8052]

If anyone sees this check out keislers elementary infinitesimal calculus, infinitesimals make the proof very easy

[@aditt.7475]

The first ten seconds of this video was the most accurate description of the chain rule that I've ever heard of.

[@arielfuxman8868]

This just shows how useful the d/dy notation is.

[@barthennin6088]

Great proof up to the point he says, "according to a previous video, this is true" (Shaking my head)

[@zbolt9able]

Excellent. So by the same logic you could turn delta u into delta z. But, the reason that doesn't work is you have no way of calculating delta z with respect to x. The reason this works is because you can calculate delta u with respect to x.

[@zbolt9able]

if 2 things have no connection to eachother, how would you calculate them with respect to eachother? for instance, how would you calculate my height with respect to your height?

[@Meverynoob]

+Akrahm Joltevskoy then how would u prove it?

[@ceryan83]

Your proof of the Chain Rule is technically incorrect. For example, what if delta u is 0 (i.e., u is a constant function of x)? In order for your argument to work, you must assume delta u is not 0, since you're dividing by delta u. Then you would have to consider the case where du/dx=0 separately.

[@Meverynoob]

what are u even asking? dy/dx = 0, dy/du = 0 and du/dx also = 0 (where u is a constant function and y is a function of u). what u are tryna talk about is delta u not being 0, which is quite obvious because u cant have a 0 in the demominator of a fraction now can you

[@ceryan83]

+Meverynoob We are assuming that u is a function of x, i.e., u = u(x). So what I'm asking is: what if u is a constant function, e.g., u(x)=1 for all real numbers x. Then delta u=0 over any sub interval. In his proof he divides by delta u, so he must logically assume that delta u is not equal to 0. So that case must be handled separately.

[@Meverynoob]

ceryan83 in that case, it'll end up with 0/0 * 0. does this not equal to 0?

[@ykkrasaoz9748]

@@Meverynoob oh dear mate lol

[@Meverynoob]

@@ykkrasaoz9748 This is me 3 years in the future and I must admit treating 0/0 as a real number is a bit embarrassing. In that case, consider y = u(x) where u(x) = 1. I just have to define u as something else. Now, dy/du is 1 but du/dx is 0. Sorry this took so long.

[@yotsubakoiwai1645]

y needs to be differentiable at u(x), not x

[@Meverynoob]

no difference.

[@Meverynoob]

in this case y is already a function of u so...

[@IvangelionTw]

@@Meverynoob so what?

[@wolframalpha8634]

If y is differentiable at u(x) it also does imply that y is differentiable at x as well , coz u(x) is a function of x , hence it's range(output) will be in terms of x(input) , and y is a function of u(x) , which implies the range of u(x) which is in terms of x turns out to be in the domain of y !! Hence we can by telling that y is differentiable at x as well!!

[@LastHopeee]

@@wolframalpha8634 we cant

[@KBineetPrasadPatro]

Thanks sir , I was founded from a long time , finally I got it . So , double thanks ☺️😊👍👌💐🎂

[@zainkhalid3049]

this video was a helpful video for me thank you. Question: (where the chain rule is use, how do we know where we use chain rule, chain rule daily life example) Thank you

[@Hobbit183]

I came up with this alternative proof: We know that: df=f'(t)*dt Now, if t itself a function of another variable x then we have that: t=t(x)=g(x). Also dt=dg (that is, an infinitesimal change in t results in an infinitesimal change in g) if we plug this into the first equation we have that: df=f'(g(x))*dg Then we divide both sides by dx: df/dx=f'(g(x))*dg/dx=f'(g(x))*g'(x) Is this correct???

[@MrZeroSugar01]

Thanks so much! I just learned the chain rule at school, and was a bit disappointed that we didn't learn a proof.. and here it is! :)

[@qualquan]

Just finding complex F'(x) by using the very useful U substitution of the "inner" x terms (function) and then finding dU by dU/dx to express everything in terms of x as a diff. coeff. Is a proof needed to justify substitution?

[@asyraafafandi1216]

what is the title of the previous video???

[@kimjongtrill2399]

The previous video is titled "Change in continuous function approaches 0"

[@aakashkarajgikar9384]

5:37 I would like an actual proof please. Is dy/dx a fraction? My calc 1 teacher/professor I have is telling me it isn't a fraction, but I am hearing wrong. So I am pretty confused.

[@shixiongong8517]

"dy/dx does not mean dy divide by dx but dy/dx is the symbol for lim(delta y /delta X) when delta X approaches 0" --my textbook

[@mohammedhisham107]

Thanks alot Khan

[@littleuniverse1430]

Can you do a proof by the definition of derivative ?

[@black_jack_meghav]

Dear sir, isn't this proof incomplete? ∆u can be zero even if ∆x isn't. This is what the textbook says.

[@nafrost2787]

4:58 I see here that you are replacing one limit here with another. Is there a name for doing that? Also what are thye conditions, that allow you to replace the limit as x -> c1, to the limit as u -> c2?

[@notsodope834]

I don't understand why is it important to consider the case when u(x) is constant function coz usually its not used. Idk

[@Sky-pg6xy]

Lol, Alsace Lorraine is also de-Frenchable

[@copernicus6420]

what a nice explanation. simply the best. you have got yourself a subscriber

[@brighttips8203]

Thnks sir

[@ElmoDaSituation]

How do you write so neatly? When i was using the scratchpad on khan academy, i found it extremely difficult to write clearly.

[@JesseMeyer]

khan uses a tablet

[@yeonhojung7185]

*ohmygod* now i get this!!

[@farruhhabibullaev5316]

It's a great description of the chain rule but the proof is not complete, You are using the theorem itself to prove.

[@drmichaelsunsschoolformath]

Not even close to a proof

[@IvangelionTw]

Why?

[@daman7387]

Bro chill it's just a KZbin video

[@Sednas]

@@daman7387 it's not a proof though and neither is it correvt

[@user-mc9pq6lc7i]

A proof should not necessarily be as clear as you wish, he just needs to show you there's no contradiction and he didn't come with a law just out of nothing..

[@martyspandex]

At last! I've been waiting for a decent proof for the chain rule for ages. Good stuff.

[@anilkumarsharma1205]

it's looking like a commutative rules of integral

[@akrocuba]

I love these vids! I always wanted to be a math/science geek, just never had the proper amount of grey matter...LOL Still, I am learning a lot....thanks for loading!

[@nileshjambhekar7699]

This is a silly question so just wanted to be clear: Why is d/dx [y{u(x))] = dy/dx?

[@paramass1802]

+Nilesh Jambhekar Hm... Your question seems to be reasonable... Actually it is not pretty clear in this video as there's some sort of confusion in the use of variables.... To understand better, y = v(u(x))=dy/dx So the modified chain rule goes as dy/dx where y=v(u(x))=(dv/du) * (du/dx) Hope this makes you understand better.. : D

[@Meverynoob]

the function is y which is one of u(x) but is also a function of x so d/dx of y as a function of u(x) is the same as dy/dx

[@praneshkulkarni4565]

Derivative of sin( sin x)

[@kostas919]

Cosx*cos(sinx)

[@MegaSairahul]

khan i wish u cud increase your pace a bit and speak like the old khan plzzzz if students can't follow some parts they will revind the video and watch it again

[@reman3000]

1.25x playback speed

[@bestillandknow2999]

Does anyone else hear "de-Frenchiate"

[@Kneecap22]

This proof is not rigorous.

[@sahandehteshami7404]

Dy/du * dx/xd = dy/dx cuz du cancels out where is my PhD

[@stevengottlieb6023]

Sal, this proof is NOT correct at all. You basically multiplied by (g(x+h)-g(x))/(g(x+h) -g(x)) which could be 0/0. This happens if in a neighborhood of x, g(x) is constant. So for h small enough g(x+h) - g(x) =0! This blows up your proof. Please remove it as it is wrong!!

[@22y4igaochenzhang8]

Bro you can't just cancel out the delta u with delta u

[@vijaynath8493]

Your voice is scary 😱😰😰😰

[@DarinBrownSJDCMath]

FAIL! This is not a proof. If you want to make little videos showing how to carry out computations, fine. Leave the proofs to the mathematicians, PLEASE.

[@kingofdice66]

I don't agree to the statement "Leave the proofs to the mathematicians". However, one must be careful and make sure he understands and give proper proofs. For example Srinivasa Ramanujan was not a professional mathematician and I'm sure that there are many examples of non professional mathematicians.

[@DarinBrownSJDCMath]

@@kingofdice66 Ramanujan was not presenting himself as a trusted authority and expert to tens of millions of people. Sal is a very smart guy, but too often he strays outside of his knowledge. Here, he is trying to present what is essentially a proof from real analysis, which is an upper-division course taken by math majors. Maybe he took real analysis, maybe he didn't. But he should know he doesn't have the mathematical background to present rigorous, abstract mathematics.

[@DarinBrownSJDCMath]

@@kingofdice66 And it's not really the fact he posted an incorrect proof that bothers me. What DOES bother me, is that AFTER the mistake has been pointed out by several people in the comments section, he has not corrected the mistake or deleted and replaced the video. Everyone makes mistakes -- especially anyone who has taught for any length of time. But if someone points out a mistake, do something to fix it. He has made several similar mistakes over the years, and not once has he fixed them. That's what I have a problem with.