Proof of the Chain Rule

[@micahchaya3909]

Sweet! Thanks for explaining that. Love proofs. Makes it easier to appreciate all the rules

[@captainamericawhyso5917]

at 4:10 why cant we also recognize that its the derivative of the composite function? Why do we need to do a variable change? Also the gear analogy slays

[@punditgi]

Unfortunately the second denominator csn be zero if an interval exists around x whrre f is constant.

[@zhangkevin6748]

Case work the rest

[@andreferreira8325]

But since we are dealing with that situation using limits solves that a problem,right?

[@Avighna]

this would imply that g'(x) is 0, right? and this would also imply that (f(g(x))' is 0, so the formula (f(g(x))' = f'(g(x)) g'(x) trivially holds true (0 = f'(g(x) * 0 ==> 0 = 0)

[@cooking60210]

Yes, the proof in this video has a gap.

[@raymondbyczko]

Very good presentation! I am a long practitioner of math, including the calculus, and use the chain rule quite instinctively. However the why evaded me me until recently. Your video helps. Two salient points helped me. (a) The chain rule is directly connected to composition functions b) The definition of a derivative function, is directly linked to a limit. Also I appreciate more the importance of differentiability and continuity. Thank you!

[@bassmaiasa1312]

The first step of the proof is not jumping out at me. In the definition of f', I would start with "g(x+h) - g(x)" as the denominator. How does h = g(x+h) - g(x)?

[@Samir-zb3xk]

As far as im aware there isnt a derivative definition where you start with g(x+h)-g(x) in the denominator. The one he used is pretty standard

[@JasurZero]

If I understood correctly, h isn't equal g(x+h)-g(x). But if h→0 => g(x+h)-g(x)→0. So we could change h→0 to g(x+h)-g(x)→0.

[@hayden6700]

When h approaches 0, then g(x+h)-g(x) approaches g(x)-g(x), which is 0. So h=h

[@JARG-Random_Guy]

No he just swiched denominators because it's multiplication.

[@arbabullah399]

Thanks sir great job 👍👍👍 I really appreciate you

[@haasjeoverkonijn6961]

Nice. But what is the font you use? I d like to use it in a report as well. Thanks!

[@dosomestuff1949]

Dude this proof doesn’t work. What if g(x+h)=g(x) on some interval centered x

[@JARG-Random_Guy]

But it's a limit. h approaches 0, it never is 0, right? Please correct me if I'm wrong!

[@JARG-Random_Guy]

Besides, of g(x+h)=g(x) you're implying that g(x)=n, so you don't need the formula just do f(n)

[@dosomestuff1949]

@@JARG-Random_Guy no its not. x close to a but not a does not imply that g(x) close to g(a) but not g(a). This causes a division by zero

[@JM_Stories1]

Best video on youtube

[@MuhammadAnas_Official]

Love it❤. 👍

[@endersteve5839]

thank you peter 🙂

[@Shubhyduby]

Thank you

[@williamvestergaard4136]

THE GOAT

[@chienbin4813]

thanks

[@SaaGT]

Thanks i understand

[@tsunningwah3471]

njh

[@JARG-Random_Guy]

njh indeed!