Why isn't abs(x) differentiable at x=0? (definition of derivative)

[@notspaso6644]

I lost it when you said "this is not 'lol', this is |0|" comedy + maths, you're the best C:

[@blackpenredpen]

= DDDDDDD

[@mannyheffley9551]

Genius

[@ashtonsmith1730]

|0|

[@somebodysomewhere9253]

*This is not lol* best quote from mathematics

[@OonHan]

Homework: Why isn't abs(x) differentiable at x=0? Me: "sites this video"

[@blackpenredpen]

That works!

[@deoxal7947]

Yes, the absolute value function's derivative is undefined at x=0, but there is an alternate definition of the derivative, which will output zero. It is called the symmetric derivative because it takes the limit from both sides simultaneously. The symmetric derivative will also output zero for certain asymptotic functions, such as 1/x^2 at x=0. An easy way of thinking about how it operates is that it takes the average of the two one-sided limits of the derivative of a function. It is for this reason that asymptotic functions will sometimes have a symmetric derivative at a discontinuity, i.e. two limits that approach positive and negative infinity cancel each other out. This is how the TI-83 and TI-84 evaluate derivatives numerically, because it speeds up computation, but it could give you an erroneous answer, so make sure you check that the original function is defined at that point before accepting its result. Graph: www.desmos.com/calculator/scrnikemaa nDeriv( command: tibasicdev.wikidot.com/nderiv Wikipedia page: en.wikipedia.org/wiki/Symmetric_derivative

[@duckymomo7935]

It has two slopes, it’s a contradiction

[@jerem33bdx]

Mi Les thanks you summed up a 10 minutes video in 1 sentence!

[@hashimabbas3977]

10:36 min video explanation in one single sentence. Thats awesome.

[@samarthsai9530]

Mi Les Well actually it has Infinite slopes , so a mega contradiction

[@yusufakn7173]

The line above me is false. Paradox yay!!..

[@DonSolaris]

*OMG!* the guy's talking to the ball!!! And i thought i saw everything on the KZbin.

[@leydigomezbravo1750]

Hi. I have been following you for 2 months and I like your videos. I really like the effort you give them and how they get to transmit. Thank you.

[@speedspeed121]

I have a good handle of my calc classes. When we have study groups, I usually teach everyone on the board the way you and Peyman present stuff. You guys inspire me to make it fun, and competitive.

[@banderfargoyl]

A function is discontinuous when the left limit doesn't equal the right limit. So you could say that a continuous function is non-differentiable when the left derivative doesn't equal the right one.

[@keerthic6925]

You are telling about the limit of f(x). Here the limit of dy/dx is taken this makes us know whether the given function is differentiable at a particular point or not

[@srinivasanveeraraagavan271]

By left and right derivative, you actually mean the left and right limits we apply to find the derivative.

[@Arthur-so2cd]

exactly :)

[@yahelck7959]

You can also find the derivative of sqrt(x^2) and get x/sqrt(x^2). At x=0 the derivative is equal to 0/0 which is undefined.

[@slowsatsuma3214]

Yahel Cohen-Kowalski Yes but what about sqrt((x^2)+1-1)

[@chronyx685]

Yahel CK use L'H rule since we can use limits here

[@FaranAiki]

Use L'H

[@notar2123]

@@chronyx685 Using L'H rule on the numerator would be searching for the derivative of sqrt(x^2), and since sqrt(x^2) is the definition of |x| you're going back to the point where you started - finding the derivative of |x|. With other words, using L'H rule is circular reasoning.

[@user-mw5tx1el9o]

absolutely hated calculus at college but really enjoyed this video. thank you

[@blackpenredpen]

: )

[@VaradMahashabde]

We could use the epsilon-omega definition of a derivative and show that (|0|)' = 0. In general then, (|x|)' = { 1 if x > 0, -1 of x < 0, 0 of x = 0}

[@scathiebaby]

Yes, that would be the function sgn(x) - Actually when you integrate sgn(x) you would end up back at abs(x) so yes, sgn(x) should be the derivative.

[@secretsquirrel4375]

Why didn’t I know about you when I was struggling through calc 2 last semester! 😭

[@blackpenredpen]

well, now you do! : )

[@Abdega]

On the bright side, now you are watching him because you want to. Not because you have to

[@mannyheffley9551]

@@Abdega well put!

[@twilight7713]

Pat pat

[@twilight7713]

It's my first semester and I knew, m lucky ig

[@weerman44]

#subbed before you hit 25k :D Your channel has grown so fast the past year! yay

[@blackpenredpen]

Yay, thank you Weerman44!!

[@gregoriousmaths266]

I can’t boast the same :( I probably hadn’t heard of calculus last year.

[@ChaosPod]

Couldn't you differentiate each piece of the function with respect to x and show they are not equal at 0 instead of taking the limit?

[@marcioamaral7511]

But we need a rigorous proof!

[@trace8617]

he “differentiated”, but used the definition of differentiability, rather than just taking the derivative

[@qtmeet]

I understand what you explained and that in this case the slope at 0 is undetermined because you can consider it having two slopes at that point, and a function can only have one y output, but I was thinking that- Maybe you could write |x| as x*sgn(x) and then the derivative is just sgn(x), and sgn(0)=0

[@qtmeet]

I dont know though, I don't have enough experience with how calculus and limits work in general

[@angelmendez-rivera351]

Not quite, no. The problem is that d(|x|)/dx = d(x sign(x))/dx = x*d(sign(x))/dx + sign(x), yet d(sign(x))/dx does not exist. Unless we allow the Dirac delta distribution to act as a derivative, in which case d(sign(x))/dx = 2δ(x), so d(|x|)/dx = 2xδ(x) + sign(x), but at x = 0, xδ(x) is not well defined?

[@jaypeethegoat]

This example was a very difficult exercise at Panhellenics Exams in Greece in 2017 for the first time in the world!

[@WolfgangBrehm]

The sign function looks like the derivative of |x| but it is defined at 0 to be 0. When we integrate sign(x) dx we get x sign(x) , which is identical to |x|. So why is the derivative of x sign(x) not equal to sign(x) ?

[@idreeskhan-zp5ey]

At 7:50 we get negative result just because of the x in the denominator,otherwise |-x|=x is always positive result?

[@alanhiguera3484]

another way to show that abs(x) is not differentiable at 0 is to define abs(x)=sqrt(x^2). differentiating abs with this defintion gives the derivative function as x/abs(x) which is not defined at x=0.

[@craig7878]

Thank you! I was looking everywhere on how understand and do this!

[@BionicMexican]

My teacher explained this, and pretty well, but seeing it all worked out using alternative def of a derivative really helped! Thanks!

[@evreng]

Also, we can write abs(x) as 2 functions. For x0, y=x. Then we can look their derivatives and as they are not same, their joint point has no derivative.

[@TheZerovirus1000]

Thanks a lot! I was soo confused when I saw a negative 1 in the limit of the abs value in my textbook

[@sabriath]

One side should be clamped in order to make the integration continuous, otherwise you end up with a disjoint when going back.....|x| is not disjointed. In this case, you can actually have both sides use "0" simultaneously, you just have to note which direction: y = {-1 if x=0+ Because if you don't, when someone were to say "what's the integration from 0 to 5" of that function, "0" cannot be evaluated, yet |x| does contain the valid positive area. If someone said to find the area from -5 to 5, you would have to break it up as -5 to 0- and 0+ to 5. etc. etc.

[@josuejaime8164]

I love your intro, Blackpenredpen Yaaaaayy!

[@blackpenredpen]

Josue Jaime yay!!! Thank you

[@sydneydwoskin4950]

somebody please get this teacher a mic

[@nsrgzh1041]

I didnt watch the video but can we say its not differentiable at 0 because we can have more than 1 line that can pass ONLY through the corner point? (goes back to the tangent line to a curve)

[@JamalAhmadMalik]

Please, give me conditions of differentiation. Is the function f(x)= arc sinx differentiable at x≥1?

[@blackpenredpen]

No, bc that's outside of the domain of arcsin(x)

[@carultch]

For real numbers, no. For complex numbers, yes.

[@grubybueno]

shouldn't it be "derivable" the title? I'm a little confused about the difference between these two concepts

[@ДанилоАнтонів]

5:35 it was too complicated!!!

[@OonHan]

As x is approaching 0 from the *negative* side, the absolute value would be equal to -x (by the definition of the absolute value).

[@ДанилоАнтонів]

it was a joke...I just can't resist when he explains such trivial things.

[@rikschaaf]

But with f(x) = |x|, does f''(0) have a value? Is it double differentiable so to speak? Let's try: f''(0) = lim_x->0 ((f'(x) - f'(0))/(x-0)) approached from the positive side: lim_x->0+ (|x|/x - (1))/x = lim_x->0+ ((1-1)/x) = lim_x->0+ 0/x, which is 0 according to wolfram alpha (0/0.0000000001 = 0, following your example) approached from the negative side: lim_x->0- (|x|/x - (-1))/x = lim_x->0- ((-1+1)/x) = lim_x->0- 0/x, which is also 0 So, since lim_x->0+ ((f'(x) - f'(0))/(x-0)) = lim_x->0- ((f'(x) - f'(0))/(x-0)) and both have an exact definition, lim_x->0 ((f'(x) - f'(0))/(x-0)) = 0, aka f''(0) = 0 Did I make any mistakes? EDIT: wolframalfa gives dirac delta function for x =/= 0 and 0 otherwise and dirac delta function = 0 if x=/=0, so basically, f''(0) = 0. I was right :D

[@SteamPunkLV]

3:14 try using L'Hopital's rule by differentiating the numerator and denominator :)

[@AaronQuitta]

ansishihi That wouldn't work because then you would get limit x->0 of sgn(x)/1 which wouldn't help you bwecause sgn(0) is undefined. If you tried to use L'Hopital again you would get limit x->0 of 0/0 which is clearly undefined.

[@SteamPunkLV]

it was a joke, because if you used L'Hopital's rule you would have to differentiate abs(x) again

[@AaronQuitta]

ansishihi Exactly, it just wasn't clear if you were unaware or joking.

[@JamalAhmadMalik]

It's sorta saying: If you can't eat a chicken, try eating a chicken.

[@SteamPunkLV]

pretty much :p

[@harshil_1.618]

That was very helpful, ty sir☺️

[@JamalAhmadMalik]

Can you please do a video on differentiation of absolute functions?

[@blackpenredpen]

I did that in the end of this vid @9:07

[@notar2123]

You can differentiate it using the definition of absolute value (|x + y*i| = sqrt(x^2 + y^2) for complex numbers, for real numbers the imaginary part is always 0 so you just end up with |x| = sqrt(x^2). The derivative of sqrt(x^2) is x/sqrt(x^2), which, interestingly enough, is the same as the signum function (sgn(x)) if x =/= 0.

[@thedanko2095]

So, If I make a tangent at x = 0, the gradient of tangent does not exist? even though a tangent can be made graphically with a zero gradient 🤔 Correct me if I am wrong.

[@redsalmon9966]

Dang... that Doraemon in the intro 😍

[@egoreremeev9969]

With another interpretation of |x| in terms of sgn(x)x, where sgn(x) returns the sign of x, or zero, it will be easier to understand, i think)

[@stewarthills9344]

The derivative of |x| is |x|/x So at x=0 the derivative is undefined

[@himanshugupta4395]

Black pen red pen I have a passage based problem from Limits and functions how can I post my doubt to you , pls help

[@blackpenredpen]

U can send it to my tweet me and I can take a look.

[@General12th]

Talk about the Weierstrass function next!

[@RAJSINGH-of9iy]

Hey i have a doubt.It is not diff at x=0 ,means we can't draw a tangent at x=0.But it seem that we cab draw a tangent which is y=0.Plzz clear my doubt

[@angelmendez-rivera351]

RAJ SINGH There are infinitely many tangent lines you can draw at x = 0, but the derivative is supposed to be unique. Hence not differentiable.

[@trace8617]

theres 270° of free space below the two functions that tangent lines can be drawn because of the sharp bend

[@19-gouthamkumarreddy58]

Come to me with a sharp pencil and paper,i will show you

[@dashboard9876]

if its limit does not exists then can it be continous ??

[@wowfmomf6126]

How about using the fact that lim when x-->0 absx over x is zero over zero, so using l'hospital's rule we derive the denominator and numerator.

[@wowfmomf6126]

The derivative of a piece wise function is piece wise function. The derivative of x is just 1. Now we can say lim when x-->0 from the right is 1 over 1 and lim when x-->0 from the left is -1/1 so left hand side does not equal right handside.

[@deepjyotibarman2499]

Sir, you from which country?

[@casdinnissen6032]

BPRP: This limit does not exist because we get 2 different values. Numberphile: Well, we'll just take the average, which is (1-1)/2 = 0 and that's the answer.

[@deserteagle3004]

Made a video about Stoke's theorem

[@WhiteDotX]

Nice video! Can you do a video explain riemman's integral vs lebesgue's intregral?

[@happyllama9900]

Could someone please explain why this is a contradiction? I'm not quite sure why it doesn't make sense to have two rates of change, especially if the point lies on the function.

[@angelmendez-rivera351]

Happy Llama It means there are infinitely many tangent lines.

[@phoebedraper3046]

so is the whole function not differentiable or just at that point?

[@notar2123]

Just at the point.

[@nicholasyap9000]

Thanks for the explanation... I have always wanted to know the derivative of that... Now that it is cleared... I have a question... Is it possible to integrate a function that is |x|?

[@RishaadKhan]

Yes! |x| has 2 cases: -x and x. Integrate each separately and the case depends on if x =

[@kye4840]

With these rules, doesn’t Lim x->0 x/x = 1? Could you explain?

[@nathanisbored]

yes it does

[@trace8617]

yes. the limit as x->anything of a constant is always equal to the constant. think about it graphically

[@wfilms6320]

thank you for existing!

[@kennethgee2004]

the slope of the tangent is 0 so it is differentiable at 0

[@angelmendez-rivera351]

Kenneth Gee No, you failed to understand. It has multiple tangent lines, so there are multiple slopes, which is why it is not differentiable.

[@ercel2010]

Then how can we say that x = nπ

[@cookiemaria780]

Does the integral exist at 0 ? If so, why? What are the differences between the criteria for differentiability and integrateability? This has been really confusing me for a while, does anyone have a moment to explain it?

[@maxsch.6555]

Could you say that the derivative of abs(x) is abs(x)/x?

[@leo0918]

Thank you so much! you saved my day.

[@morgengabe1]

Been hitting The gym hard lately

[@dildobaggins2759]

why isnt 0 included in the piecewise function? You have greater than x values and less than x values but you havent done a calculation with I0I/0

[@MrQwefty]

The function is continuous but its derivative is not continuous

[@vahidy2002]

Wonder if we get any application of this kind of function .

[@hhho192]

0:03 what the hell was that.....a multi coloured Doreman..?

[@tamoghnamaitra9901]

Awesome 😊. Really helped

[@noahniederklein8081]

The absolute value of 0 isn't a joke guys, stop laughing

[@kenichimori8533]

f(3) + f(2) = f(1)

[@sardarbekomurbekov1030]

Nice proof

[@blackpenredpen]

: )

[@OtherTheDave]

What’s wrong with saying “f'(0) = [-1, 1]”?

[@angelmendez-rivera351]

OtherTheDave “It’s not a number”.

[@angelmendez-rivera351]

Truth be told, there is nothing wrong with it, it’s just that in calculus, they are really stingy about things

[@bootcamp6628]

But |x^odd power| except 1 is differentiable at x=0...y this??

[@adi7828]

IIT be like : Hmmm, I bet students didn't know this . Let's make an advance question 😈😈😈

[@roger72715]

Thanks..it helped!

[@blackpenredpen]

Yay!!!

[@BedrockBlocker]

I know this is dangerous territory, but couldn't you say the limit has two values?

[@angelmendez-rivera351]

David Birkenmayer No. By definition, limits are single-values.

[@purim_sakamoto]

ほむほむ negのとこでは-1に傾いてて、posのとこでは1に傾いてるけど 0のとこでは別に水平でもなんでもなく、どっち向いてるかわかんない！と うううーーーんもにょる　これリーマン系だからそうであって、他の系では定義されるんじゃないかってすっごい気になる イライラしちゃうわァ･･･

[@zackiz]

How can the function be continuous at x=0 if the left and right limit doesn't agree? Isn't the equality of the left and right limit a condition for continuity?

[@davidgould9431]

The function f(x) = |x| is continuous at x=0 because both limits (from -ve and from +ve) of f(x) tend to 0. It's the derivative, f'(x), that tends to different values from each side, namely -1 from the left and +1 from the right.

[@OonHan]

Hi!

[@raghavvishwanath6570]

#subbed since before you even hit 50k

[@blackpenredpen]

yay, thank you!!

[@AhnafAbdullah]

Same

[@animeonguitar3850]

he used blue pen too

[@alanturingtesla]

Well, in machine learning you set it to be whatever, it does not really matter. Ah, engineers...

[@naseebullah5476]

great job

[@raghavvishwanath6570]

This is not lol😂😂

[@blackpenredpen]

: )

[@leemonsyzure6010]

so you could just write the derivative of abs(x) as abs(x)/x couldn't you?

[@emiliadaria]

Math is the best, isn't it? 😀 #YAY

[@blackpenredpen]

Yay!

[@VonUndZuCaesar]

Does the 2nd gradient exists at x=0 for the absolute function?

[@angelmendez-rivera351]

TheCaesar2011 no

[@Dalton1294]

#subbed I enjoy watching your videos

[@Владимир-н8я1г]

Thanks

[@snbeast9545]

What'd happen if you used the actual definition of absolute value (which is the radius of the polar form)?

[@notar2123]

As I've just said on another comment, since this is dealing with real numbers, you'd get |x| = sqrt(x^2). If you differentiate sqrt(x^2) you get x/sqrt(x^2), so obviously sqrt(x^2) has to be not equal to 0, which again means x cannot be 0. You get the same thing basically.

[@anthonySung__]

The only thing I learnt from this video: |0| is not lol

[@giovannimariotte4993]

Very interesting

[@ListentoGallegos]

so the derivative of abs(x) exists but just not at x=0?

[@debrajbanerjee9276]

So now you will change your profile picture by doramon

[@AlgyCuber]

hi already #subbed like 3 months ago

[@blackpenredpen]

Algy Cuber yay!!!!

[@friendlydragon8999]

I just love calculus.

[@randomtraveller7083]

Dragon

[@MetallicDETHmaiden]

its not dable, isnt it xD

[@وريانهاد]

Is it differentiable at x=3 ??? I think yes

[@ShyamSingh-ui4ih]

Good

[@SeriousApache]

You still using finger

[@mathematicstvc6000]

Hi

[@livebyfaithnotbysight9134]

Thank you!!! :)

[@Bluedragon2513]

It's a 5 guys

[@tranquillities1205]

first of all, |X| doesn't not equal -X for values of X

[@angelmendez-rivera351]

l l Holy shit, you are so stupid. It’s pretty clear you haven’t picked up a book. if x < 0, then |x| does indeed equal -x, because if x < 0, then -x > 0, which means -x is positive. Geez. “But convince yourself of these facts before making another video.” How about you get a degree and convince yourself of any facts st all before you ever write another comment,

[@forgedwithsteel]

my man you are confused, multiplying an x that is less than 0 by -1 gives you the same value but positive. which is the answer to |x| of any negative number. |x| is just the distance from the origin on a graph lol

[@kishorbhushan8292]

I love doraemon