Thank you for this. I wish I had this 5, if not 10 years ago. I find that most Calculus textbooks define the concept of a limit, show a few examples and then go over derivatives and then stop explicitly linking the two again. It then tends to be found at the beginning of integration and the same story occurs. I think that more should be shown with limits as they really are the foundation that Calculus is based off of.
@@jadegrace1312 But other functions could be constant over some interval around a. This is a well known flaw on the proof as presented, which otherwise is very nicely done.
@@jadegrace1312 The problem is that the totally correct proof is a little more complicated in order to avoid that division. I hope our presenter shows us the revised proof with his usual clarity. 😃
There’s nothing wrong with the proof. Every derivative definition has a “divide by zero”. But this division is inside a limit in an indeterminate form ( 0 / 0), so we can use limit properties to evaluate the limit.
Hi, A rapid way for showing the derivative of the ratio. But may be we can exibit this from the product rule: We know that (fg)’ = f’g+fg’, that tells us that f’g = (fg)’ - fg’, and then f’ = ((fg)’ - fg’ )/g . Let u = fg, then f = u/g, and then (u/g)’ = (u’ - ug’/g)/g . Let’s multiply up and down by g : (u/g)’ = ( u’g-ug’)/g² For fun: 1 "and so on and so forth", 12 "go ahead and ...", including 9 "let's go ahead and ...", inluding 2 "let's go ahead and do that", 6 "great", and the usual "and that's a good place to stop".
2:20: for g(a) = c, we apply the product rule and get (fg)'(a) = f'(a) c + f(a) 0, as long as f is not infinity then this results in a trivial expression cf'(a)
Finally something I can understand lol. Taking a Calc 1 type of class with stat for my BA bSc which has kind of nothing to do with math in my opinion but hey, they force you to take it to ''show your brain cells are working'' from my understanding lol... But I like to watch these cool math videos to understand better and hopefully pass my exam
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120, what is the value of s(3m)? Can someone please help me with this? It came in India. PRMO 2019 25th August Paper
Lim x->a f(x)=f(a) is something basic that you probably know given f cont. at a. Looking at his original definition of g’(f(a)) involving y, it uses lim y -> f(a). We can replace the y with anything and adjust the lim something -> something else once the y placeholder still approaches f(a). Therefore, if he replaces y with f(x), and we know lim x-> a f(x) = f(a) , thus that lim something -> something else is lim x-> a, so then the y placeholder (which is f(x)) becomes f(a) as required. I might even be ‘over’ explaining this as it’s really a simple property of limits, but I suspect you’re overthinking it or getting confused among the rest of the contents of the video. Hope it helps.
I would start by trying to roll the function up into smaller bits maybe and then trying to see if there is any element you can apply in order to give some sort of an equation to the problems... hopeit helps
14:21
That’s a good place to stop
When will you do the AMA?
Thank you for this. I wish I had this 5, if not 10 years ago. I find that most Calculus textbooks define the concept of a limit, show a few examples and then go over derivatives and then stop explicitly linking the two again. It then tends to be found at the beginning of integration and the same story occurs. I think that more should be shown with limits as they really are the foundation that Calculus is based off of.
May not have been a good idea to use a as your variable of choice here
I've always wanted a proof of this since high school! I'm 29 now and you made my wish come true finally :D
For the chain rule if f(x) = f(a) for some neighborhood of a, then the proof fails because of division by zero
I don't think that's true because if you use f(x)=x then the proof works fine if you follow the same steps
@@jadegrace1312 But other functions could be constant over some interval around a. This is a well known flaw on the proof as presented, which otherwise is very nicely done.
@@punditgi That makes sense. Very interesting.
@@jadegrace1312 The problem is that the totally correct proof is a little more complicated in order to avoid that division. I hope our presenter shows us the revised proof with his usual clarity. 😃
There’s nothing wrong with the proof. Every derivative definition has a “divide by zero”. But this division is inside a limit in an indeterminate form ( 0 / 0), so we can use limit properties to evaluate the limit.
Hi,
A rapid way for showing the derivative of the ratio. But may be we can exibit this from the product rule:
We know that (fg)’ = f’g+fg’, that tells us that f’g = (fg)’ - fg’, and then f’ = ((fg)’ - fg’ )/g .
Let u = fg, then f = u/g, and then (u/g)’ = (u’ - ug’/g)/g .
Let’s multiply up and down by g : (u/g)’ = ( u’g-ug’)/g²
For fun:
1 "and so on and so forth",
12 "go ahead and ...", including 9 "let's go ahead and ...", inluding 2 "let's go ahead and do that",
6 "great",
and the usual "and that's a good place to stop".
Cool, suddenly the derivative rules become so easy and understandable
2:20: for g(a) = c, we apply the product rule and get (fg)'(a) = f'(a) c + f(a) 0, as long as f is not infinity then this results in a trivial expression cf'(a)
I am struggling here with finding derivates in points of the inverse functions and this guy makes this all look so easy LOL
mike i can't find the playlist on real analysis anymore .............. the main reason i became your virtual student
Finally something I can understand lol. Taking a Calc 1 type of class with stat for my BA bSc which has kind of nothing to do with math in my opinion but hey, they force you to take it to ''show your brain cells are working'' from my understanding lol... But I like to watch these cool math videos to understand better and hopefully pass my exam
Best of luck. Don’t get overly taken up with real analysis...It’s a bit much for only calc one but definitely watch if genuinely interested.
this course style reminds me of rod haggarty textbook
excellent stuff the way you teach most people skip his stuff cause they confused but you convey it an excellent manor
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120,
what is the value of s(3m)?
Can someone please help me with this?
It came in India. PRMO 2019 25th August Paper
Could you explain the compositional limit part at 8:51?
Lim x->a f(x)=f(a) is something basic that you probably know given f cont. at a. Looking at his original definition of g’(f(a)) involving y, it uses lim y -> f(a). We can replace the y with anything and adjust the lim something -> something else once the y placeholder still approaches f(a). Therefore, if he replaces y with f(x), and we know lim x-> a f(x) = f(a) , thus that lim something -> something else is lim x-> a, so then the y placeholder (which is f(x)) becomes f(a) as required. I might even be ‘over’ explaining this as it’s really a simple property of limits, but I suspect you’re overthinking it or getting confused among the rest of the contents of the video. Hope it helps.
🔥🔥🔥
Im having trouble with this problem can you please help me. Find the function f(a(g)) where a(69) = 80085 and f(420) = 911
I would start by trying to roll the function up into smaller bits maybe and then trying to see if there is any element you can apply in order to give some sort of an equation to the problems... hopeit helps
Why does this function look like a joke?
Oh shit, it is
here is a vid that might help! ;) czcams.com/video/6_b7RDuLwcI/video.html
for example [(80085/69)*x]^k=911, there is k=ln 911/ln(80085*420/69), (we take the linear function only for x>0, there exists ln function)