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)
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!
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)?
Sweet! Thanks for explaining that. Love proofs. Makes it easier to appreciate all the rules
Unfortunately the second denominator csn be zero if an interval exists around x whrre f is constant.
Case work the rest
But since we are dealing with that situation using limits solves that a problem,right?
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)
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!
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)?
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
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.
Thanks sir great job 👍👍👍 I really appreciate you
Nice. But what is the font you use? I d like to use it in a report as well. Thanks!
thank you peter 🙂
Thank you
Love it❤. 👍
thanks
THE GOAT
Best video on youtube
Thanks i understand
njh