If you think of the partial derivative as the component of the gradient vector, chain rule just pops out by definition of how covariant vector components transform which is pretty neat. Not sure if it also works for chen lu.
At 13:52 I am a tiny bit confused. So you are taking the limit as h->0 of the error term of h squiggle (which has h in the def and would to zero as h goes to zero. Though with that logic, wouldn’t you have had to move the limit inside of the error function? And to do that, we would have to prove the the error function is continuous right? I just don’t remember if that was done or it is was super easy to see why
Everett Meekins when I had to do this proof in my Real Analysis test, yes, we had to prove the continuity of the function. It's a really long and dull proof
Hey Dr. P. New subscriber. Love your vids. One suggestion on this one is that the board doesn't fill the screen. Makes it harder to follow on a phone screen. No probs. I can grab the tablet if needed. 😎
I've seen a different proof that defines the two functions : given y_0=g(x_0) (and calculating the derivative of f composed with g at x_0, with g differentiable at x_0 and f differentiable at y_0) F(y)={f'(y_0) when y=y_0, (f(y)-f(y_0))/(y-y_0) otherwise. G(x)={g'(x_0) when x=x_0, (g(x)-g(x_0))/(x-x_0) otherwise. Then shows that (f(g(x))-f(g(x_0)))/(x-x_0) = F(g(x)) G(x) since both F, g and G are continuous at the appropriate points (which is because f and g are differentiable at y_0 and x_0 respectively), when taking the limit we get F(g(x_0)) G(x_0), which is by definition f'(g(x_0)) g'(x_0)
When I had to prove the chain rule, I never really understood why it worked (it doesn't help that the schoolbook I had has got a confusing proof). I knew that I needed a second variable (k in that case) which is dependent on h but I never came into me where that k came from. The punchline for proof of the chain rule is to define k as v(x + h) - v(x) (numerator of the inner differential quotient) and make the outer limit dependent on k instead of h. Due to the definition, if h -> 0 then k -> 0 too because v(x + 0) - v(x) = v(x) - v(x) = 0. That way, you can define as u(v(x + h)) - u(v(x)) as u(v(x) + k) - (u(v(x)) as shown in this video and derive the outer function. Of course, your proof goes even more technical but the most important step in the chain rule is to define k as v(x + h) - v(x).
Leithold says the function F in delta u terms, might be continuos at zero.. but Why? Derivation`s condition? I am a engineer , two weeks stuck wthis! Greetings from Perù!
ONE potential problem..... we agreed Sigma is some junk..... but if it depends on h or h « squiggle »(I prefer to call a tilde a tilde, unless it is a Waltzing MaTilde) suppose Sigma blows up? Can we show sigma is stationary (constant) or only increases in such a way that it’s product with f(x+h) - f(x) still goes to zero?
I have a faster proof. It involves what you already did with factorizing out f'(x) out of d/dx g(f(x). the 2 terms you are left with is (g(f(x+h)-g(f(x))/f(x+h)-f(x) * f'(x). I will focus on the first term. substitute f(x+h)-f(x) =u and f(x)= k You'll be left with lim as u--> 0 of (g(k+u) - g(k))/u * f'(x) Which is just g'(k) * f'(x) = g'(f(x) * f'(x). Bada bing bada boom QED baby.
Doesn’t that circumvent the limiting process as part of the definition of the differential? So.... pretending the numerators and denominators of differentials are separable is quackery!!!! (EverybodyDUCK!!!)
If you think of the partial derivative as the component of the gradient vector, chain rule just pops out by definition of how covariant vector components transform which is pretty neat. Not sure if it also works for chen lu.
Dr peyam i took complex analysis this semester because you motivated me so much to take it all last year!
Chen Lu, not Chen Lou 😅
Chen Lew?
Chen Lieu
en.m.wikipedia.org/wiki/Chen_Lu_(figure_skater)
@@koenth2359 Imagine every calling her chain rule now lmao. She will be so confused.
Chen Liu.
Simple, efficace ; merci docteur, vous mériteriez vraiment plus de visibilité sur la plateforme ! (PS : Chêne Lou ça marche aussi)
Nn c'est tchene lou
Ou peut-être chien-loup
I have only recently started watching your videos. I love them.
Clarísimo, Fantástico...Gracias, Dr. Peyam.
This is one of my favourite episodes of the Dr Peyam show. I feel the Chen Lu coursing through my veins now 😁😁
Time for the HD Chen Lu!
Chen Lu *does* sound powerful... 🤔 Maybe I'm gonna use it in a future video.. :D
Peyam, 100k is getting so close for you. I have no idea how you didn’t reach it years ago, but atleast your still our little secret lol!
Awwww thanks so much!
Why is there a junk term in the g'(y) ?
At 13:52 I am a tiny bit confused. So you are taking the limit as h->0 of the error term of h squiggle (which has h in the def and would to zero as h goes to zero. Though with that logic, wouldn’t you have had to move the limit inside of the error function? And to do that, we would have to prove the the error function is continuous right? I just don’t remember if that was done or it is was super easy to see why
Everett Meekins when I had to do this proof in my Real Analysis test, yes, we had to prove the continuity of the function. It's a really long and dull proof
paulo ed machado oh okay. I just started taking real analysis the semester (only have had 2 classes). But rip tho
Hey Dr. P.
New subscriber. Love your vids. One suggestion on this one is that the board doesn't fill the screen. Makes it harder to follow on a phone screen. No probs. I can grab the tablet if needed. 😎
Same here Adam. Difficult to read on my phone. Beautiful proof though.
Rotate sideways :)
6:52 are we allowed to multiple out by h squiggle if it's zero? Would the zero denominator issue?
Precisely what I wanted to learn today!
Was your lecturer Chinese because in Mandarin （律 =lu) means rule and I'm assuming 'Chen' is chain pronounced with an accent?
Ooooh I had no idea that Lu literally means rule!!!
I've seen a different proof that defines the two functions :
given y_0=g(x_0) (and calculating the derivative of f composed with g at x_0, with g differentiable at x_0 and f differentiable at y_0)
F(y)={f'(y_0) when y=y_0, (f(y)-f(y_0))/(y-y_0) otherwise.
G(x)={g'(x_0) when x=x_0, (g(x)-g(x_0))/(x-x_0) otherwise.
Then shows that (f(g(x))-f(g(x_0)))/(x-x_0) = F(g(x)) G(x)
since both F, g and G are continuous at the appropriate points (which is because f and g are differentiable at y_0 and x_0 respectively), when taking the limit we get F(g(x_0)) G(x_0), which is by definition f'(g(x_0)) g'(x_0)
When I had to prove the chain rule, I never really understood why it worked (it doesn't help that the schoolbook I had has got a confusing proof). I knew that I needed a second variable (k in that case) which is dependent on h but I never came into me where that k came from.
The punchline for proof of the chain rule is to define k as v(x + h) - v(x) (numerator of the inner differential quotient) and make the outer limit dependent on k instead of h. Due to the definition, if h -> 0 then k -> 0 too because v(x + 0) - v(x) = v(x) - v(x) = 0. That way, you can define as u(v(x + h)) - u(v(x)) as u(v(x) + k) - (u(v(x)) as shown in this video and derive the outer function.
Of course, your proof goes even more technical but the most important step in the chain rule is to define k as v(x + h) - v(x).
Time to understand the rule that the teachers made us memorise it !!!
Leithold says the function F in delta u terms, might be continuos at zero.. but Why? Derivation`s condition? I am a engineer , two weeks stuck wthis! Greetings from Perù!
Please proof the theorem of chain rule for integral calculus
ONE potential problem..... we agreed Sigma is some junk..... but if it depends on h or h « squiggle »(I prefer to call a tilde a tilde, unless it is a Waltzing MaTilde) suppose Sigma blows up? Can we show sigma is stationary (constant) or only increases in such a way that it’s product with f(x+h) - f(x) still goes to zero?
Is there an easier way?
Nope
That was beautiful
Awesome
Why sigma -> 0 when h ->0? Sigma isn't a continuous function when f(x+h) - f(x) = 0
So what actually changed when you replaced f(x+h)-f(x) with h squiggle? You just substituted it back later... Sorry if I'm missing something obvious
_Use the Chen Lu!!!_
...and don't forget to wash your hands.
I skipped to the end just to hear the story again
You are so crazy Blackpenredpen and you ! I understand now. My best friend ! It's good all that the morning. Thanks.
Chen lu can also be the Latin expression of a Chinese girl’s name。。。。。。
Use the Chen Lu!!!!
I have a faster proof.
It involves what you already did with factorizing out f'(x) out of d/dx g(f(x).
the 2 terms you are left with is (g(f(x+h)-g(f(x))/f(x+h)-f(x) * f'(x).
I will focus on the first term.
substitute f(x+h)-f(x) =u
and f(x)= k
You'll be left with lim as u--> 0 of (g(k+u) - g(k))/u * f'(x)
Which is just g'(k) * f'(x) = g'(f(x) * f'(x).
Bada bing bada boom QED baby.
Simplest proof: dy/dx =dy/dx * du/du = dy/du * du/dx
HelloItsMe
Love it!
Doesn’t that circumvent the limiting process as part of the definition of the differential? So.... pretending the numerators and denominators of differentials are separable is quackery!!!! (EverybodyDUCK!!!)
I am very interesting but. I am not see anything because it too small
I was an awsome man, then I was
Chung li from fortnite
Quit ur job kid im smarter
This seems like non-standard analysis, with the sigma term being an infinitesimal.