Proof: Differentiability implies continuity | Derivative rules | AP Calculus AB | Khan Academy

[@arunkhanna7210]

Differentiability implies Khantinuity!!!

[@paranjoykb]

he he yes

[@jasonmiller8407]

I khancur

[@randomeggthatworksforthefb7172]

This is super useful actually.

[@NewWesternFront]

can't slant the Khant (only the tan line) ayyyyy

[@225vikrant3]

😂😂nice

[@emresarac2100]

I wish my teacher taught that like this way. He just goes too fast that we cannot follow what he's doing xd. Thanks a lot Khan Academy!

[@giack6235]

Hello and thank you for the video, but I think there's an error in 9:44: If you have lim (x -> c) (f(x) - f(c)) = 0, you can conclude that this is equivalent to: lim(x -> c) f(x) = lim(x -> c) f(c) only if you know that the single limits exist, but you don't know yet if lim(x -> c) f(x) actually exist!

[@cansucorbac1514]

"The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn." :)))) Canım.

[@MyAsdfqwe]

canın.

[@esraeren00]

actually calculus course really require knowling the proof! ın the exams 75 percent of questions are measued our ability to prove them

[@booguy2636]

Why do you start the proof with lim x→c (f(x) - f(c))?

[@TU7OV]

Because to show that lim x->c f(x) = f(c) you have to show that lim x->c f(x) - lim x->c f(c) =0 or lim x->c (f(x) - f(c)) = 0, this follows from the limit laws.

[@aanyaakhandelwal1778]

Thank you very much!!! So good!!!

[@mehulkathiriya1737]

awesome...I like it even more than iit lecture.short and simple....

[@serve932]

I need to learn this 4 times for it to get inside my head

[@benwinstanleymusic]

Thank you very much Sir!

[@arunkumar4u]

Why you took lim( f(x)-f(c)) ?? x>c

[@booguy2636]

I'm wondering the same thing.

[@SadikulIslam-cm8wo]

Really awesome

[@rajeshghosh8737]

Nice explanation.........thanks.

[@Hakahah_o]

Thank you

[@jamessamuel1255]

Bloody brilliant

[@cyvex1281]

Short and simple

[@rainerwahnsinn3262]

You are only allowed to split the limit, if both parts are *convergent*!

[@manivannan235]

Yes it is indeed convergent😊

[@stickmanbattle997]

This video only proves limx->c(f(x)) = f(c), but the most important info if the lim is not continuous at that point it means is not differentiable.

[@sujitbaruah4536]

Nice lecture

[@wagsman9999]

Excellent. I suppose you can assume f(c) exists, because if it didn't the derivative would fail.

[@Lark-um8hv]

Can't have the slope at a point if the point doesn't exist

[@ghassanayyad9766]

thank you so much

[@aanyaakhandelwal1778]

Yes, you are very corrrect!

[@randomeggthatworksforthefb7172]

Plus eggs are tasty

[@hassanakhtar7874]

Œ

[@lizmurith9861]

Why do you multiply and divide by (x-c)? Why are we changing the original formula?

[@MrCanadianplayer]

Because it's a constant Liz! (x-c) over (x-c) is just like saying 1/1=1

[@lizmurith9861]

Manchester United Videos Thanks!!!

[@MrCanadianplayer]

Liz Menezes no problem lolll

[@MrCanadianplayer]

Liz Menezes I actually have a test on this stuff tomorrow haha can you answer my question now? Why is Lim x approaching c of x-c =0? Hope you can help me out Liz! Matan

[@MrCanadianplayer]

Liz Menezes so F(c) =0?

[@jaswanthramesh2401]

can f'(C) be infinity?(happens for a vertical straight line) , if it is infinity, then 0*infinity becomes an indeterminate form . Then we wont get this.

[@jakubsmyk2251]

if we assume differentiability of function f at point C then f'(C) is finite

[@akames.529]

7:24 where does (x-c) come from?

[@Lark-um8hv]

w = (q/q)w = (w*w)/q = w × (w/q) Try it with constants

[@DynV]

TLDR 10:56

[@CMDF08]

I think I missed something here. Why can we say that the limit of f(c) as x approaches c is equal to f(c)? I don't see where we proved that anywhere. I don't understand where it comes from. I must have missed some detail. I understand how we proved that the limit of f(x)-f(c) as x approaches c = 0, but if our proof is using the fact that f is continuous at c, isn't that circular logic? Or does the limit of f(x)-f(c) as x approaches c = 0 somehow require f to be continuous at c? I'm so lost :(

[@jeanbadgenes8460]

yeah, he's wrong, they are not equal, thats why is discontinuous

[@tansu1499]

He's right because f(c) is a number. that's one of the rules of limit , when you want to find limit of a number the answer "always" is that number. so here limit of f(c) always equals to f(c). If you didn't get it just watch a video about rules of finding limit.

[@CMDF08]

​@@tansu1499 I think I understand now. Since f(c) evaluates to the value at c regardless of whether the function is continuous or not, then f(c) would certainly be some unknown number. My only issue now is what if f(x) were undefined at c? In that case, we could not say f(c) is a number, right? Aren't we also depending on the idea that a function that is differentiable at c is also defined at c? Perhaps there is another, independent proof that establishes this though, that I've forgotten as I haven't looked back at this for a while.

[@CMDF08]

Now that I look at the formula for differentiation, actually, perhaps it is trivial to say that if f(c) were undefined at c. We would not be able to apply that formula otherwise, and therefore it must be defined at c if it is differentiable at c.

[@tansu1499]

@@CMDF08 exactly. We know that it's differentiable at x=c so it is defined at this point and then we use these steps to prove that it is also continuous at this point.

[@Dyslexic-Artist-Theory-on-Time]

Can we think of this as a process over a period of time?

[@Lark-um8hv]

Sure, if x represents time why not?

[@drallisimo34]

cool!!!

[@aragonification]

But if differentiability at a point means that the derivative of that point exists, why the first example from the right side is not continuous?

[@wolframalpha8634]

As per the definition of continuity lim as x approaches point c from the right as well as the left and also if the function is defined at that point C then it is equal to f(c), whereas the function at point c is defined at some other value of f(x) , i mean a removable discontinuity... instead if we were to look at the actual position of f(c)(marked with an empty circle) the left hand limit and the right hand limit exists at that point but is not equal to f(c) cause it is not defined at that point!! Hence the function is neither continuous nor differentiable at that point

[@imdoug]

math god

[@puspitapaul725]

please explain the differentiability of |log|x||

[@renayachan5714]

Does it mean that if a function aint continuous at a number it is also not differentiable at that number?

[@montrast0705]

Yes. A implies B is equvialent to notB implies notA. In your case A being function is differentiable and B being function is continuous.

[@Mooreeezy]

well differentiability implies continuity. however continuity does not necessarily imply differentiability. Which means that if the function is not continuous then that function could sill be differentiable.Because differentiation is basically the process of finding the derivative.And a derivative is a means whereby we are able to find the slope of a non linear function at any point on that function. Continuity suggests that a function is continuous. Does a function need to be continuous in order to be able to use the same derivative to find any slope on it? I don't believe so. So, in my mind that means that continuity does not imply differentiation. which I think means a function does not have to be continuous to be differentiable. *UPDATE* thought about it some more. If you have a gap and you use the derivative to find the slope of the function with the gap, if you plug an x value into the derivative and it ends up in the gap, that would suggest that the derivative does not work for that specific point on the graph. which means you would have to use a different derivative. which means the graph wouldn't be differentiable at that number. So I guess I really don't know.

[@Muhammed_English314]

@@Mooreeezy If you know a little bit about logic (like me) you will conclude that the answer is yes because "discontinuity implies undifferentiability" is the contrapositive of "differentiability implies continuity"

[@vvr311]

what can you say about differentiability of y = x^(1/3) and y = x^(2/3)?????

[@Muhammed_English314]

they are both differentiable on the real numbers except at x=0 1) y'=1/(3cube root of x squared) so if you look at the domain of the derivative function it's R/{0} 2) just the same but without the square

[@majorgeneralrahul6298]

Fun Fact :- Even if you assume it's NOT differentiable you'll get the same answer if you use this method. :D

[@kamitube1059]

I lost at 7:10 , can anybody please help me? why is he multiplying (x-c) at top and bottom?

[@lebongarcon3390]

he multiplies it top and bottom so it is equal to the original limit function. He is doing this to connect the derivative function to the limit function

[@paulpandig9325]

try to arrange the video in order that means first proof and then sum

[@yashuppot3214]

The audio...

[@ian.ambrose]

I know. It's great.

[@ahmedalsherbini276]

thumb up

[@prajwol_poudel]

writing this in my exams aint gonna score me good marks

[@Muhammed_English314]

learning is not for grades it's for science and fun.

[@TheSharingarMaster]

Thank you