Does Derivative Have to be Continuous? (feat. x^2sin(1/x))

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  • Опубликовано: 24 дек 2024

Комментарии • 44

  • @LetsSolveMathProblems
    @LetsSolveMathProblems  6 лет назад +8

    For the example at the end concerning infinitely many discontinuity of the derivative, visualize a function that resembles the behavior of the near-zero part of x^2sin(1/x) at integer points. Copy-and-pasting a part of x^2sin(1/x), I realized after making the video, does not quite work because it is an odd function and there will be problem of whether the function will be differentiable at the boundary between two equal parts (even if we change the interval we copy-and-paste). However, the intuition still stands.

    • @laugernberg4817
      @laugernberg4817 6 лет назад +1

      yes, and specifically it is possible to let these intervals "(-2,2)" be smaller and smaller!!. Ex copy it onto [1/2,1], and [1/3,1/2), and [1/4,1/3) and so on. Then the derivative has *infinitely* many discontinous points on a *bounded* interval (0,1]!

    • @shagunagrawal2003
      @shagunagrawal2003 4 года назад

      How does this cutting-pasting concept work to show that f' is discts. at infinitely many points on the real line? I didn't get it at all. :(

  • @CyrusSaiyed
    @CyrusSaiyed 8 месяцев назад +1

    Please make a video on (x - 3)(x - 1)|(x - 3)(x - 1)|

  • @moeberry8226
    @moeberry8226 2 года назад +1

    Hands down the best Video of reql Analysis on RUclips. LetssolveMathProblems is the master of Mathematics.

  • @divisix024
    @divisix024 4 года назад +4

    This is the standard example in the calculus class I took

  • @Barman1904
    @Barman1904 3 года назад

    At 5:50 ....acc to the definition of derivative we are getting a finite value of derivative of the function at x=0 but by the usual method we are not getting so as limx→0(cos 1/x ) isnt defined .....why such thing is happening????

  • @guycohen4403
    @guycohen4403 8 месяцев назад

    How will the function change if instead x²sin(1/x) it will be x²sin(1/x²)?

  • @gauravjindal6148
    @gauravjindal6148 4 года назад +1

    Please tell very important here if going through first derivative principle f'(0+)=0 but differentiating the function gives that f'(0+) is osscilating,, can we say that first principle dont give us derivative at f(0+)

  • @lucas29476
    @lucas29476 6 лет назад +10

    The cool thing is that if you change the x^2 to x^3, then it is not a counterexample. I’m still trying to find an intuitive reason for this difference.

    • @iabervon
      @iabervon 6 лет назад

      If you look at the graph of the counterexample, you can see that the derivative has magnitude 1 every time it crosses 0 (except at 0), and opposite signs each time. The function stays continuous by going less far each time as you approach 0, but this doesn't help the derivative. With x^3, the closer crossings are also flatter, so the derivative can be continuous.

    • @stephenbeck7222
      @stephenbeck7222 6 лет назад +3

      Yep. I found it helpful a while back to consider the chain of functions, starting with f1(x)=sin(1/x) like he does in the video, and as discussed f1(x) does not have a limit as x approaches 0. Then you have f2(x) = x*sin(1/x), which does have a limit at x = 0 (from here we can make the piecewise function with f(0)=0 to make it continuous) but does not have a derivative. Next is f3(x) = x^2*sin(1/x) which has a first derivative that itself is not continuous. And continuing, f4(x) = x^3*sin(1/x), a function with a discontinuous 2nd derivative. And we could keep going on.

    • @shubham1999
      @shubham1999 6 лет назад

      Cool try it.

  • @rabeeashaheen6778
    @rabeeashaheen6778 4 года назад

    What about square root of x into sin(1/x)?? Please help me out.

  • @quahntasy
    @quahntasy 6 лет назад +4

    Nice video. Loved it!

  • @yoavshati
    @yoavshati 6 лет назад +1

    Couldn't you use {x at x=/=0 and 0 at x=0} as a counterexample?
    The derivative is 1 for all x but 0, and it's 0 at x=0

    • @ais4185
      @ais4185 6 лет назад +1

      From the definition of the derivative (like he did in the video), the derivative at x=0 would actually be f'(0)=lim h->0 ( f(0+h)-f(0) )/ (0+h-0)= lim h->0 (h-0)/h=1

    • @LetsSolveMathProblems
      @LetsSolveMathProblems  6 лет назад +2

      We cannot because the derivative is 1 (NOT 0) at x = 0 as well. Note that f'(0) = lim as x->0 of (f(x)-f(0))/(x-0) = lim as x->0 of x/x = 1.

    • @RGP_Maths
      @RGP_Maths 6 лет назад

      Yoav Shati Your function is no different from f(x)=x, therefore it's obvious that the derivative is 1 everywhere, as shown more formally in other comments.

    • @Muhammed_English314
      @Muhammed_English314 3 года назад

      That's a common mistake in peace wise functions, when you have a single point defined like f(a)=b people assume that the function is constant or a horizontal line at that point (a,b).
      I made this mistake a few times and it really feels satisfying when you find out why what you did is wrong and understand the subject more.

  • @BB-fp9ce
    @BB-fp9ce 4 года назад

    tell pl me why are you looking at limx->0 , but nio limx->0- and x->0+ ? lim(sin1/x) x->0+ = 0

  • @ashudinijr2911
    @ashudinijr2911 5 лет назад

    f(x) = { x sin 1/x, x is not equal to zero. { 0, x=0
    Is this function continuous every where or continuous only at x=0 plz tell me

  • @fpv10
    @fpv10 3 года назад +6

    영언데 한국어처럼 들리네ㅋㅋㅋ

  • @nadineahmed1249
    @nadineahmed1249 4 года назад +1

    Thank you so so so so much!! This video was so helpful!!!!

  • @matteosanturi1643
    @matteosanturi1643 6 лет назад +2

    I'm ok that you showed an example, but next time could you prove it by using theorems? Thanks :)

    • @Koisheep
      @Koisheep 6 лет назад

      Some functions can be continuous, differentiable and have continuous derivative everywhere, whereas other functions (like this) don't. You can't prove a certain property is true/false *sometimes*, only that it's ALWAYS true/false. For the *sometimes*, all you can find are examples where it's true and examples where it's false.

    • @LetsSolveMathProblems
      @LetsSolveMathProblems  6 лет назад +15

      Using a counterexample to prove that something cannot be true is completely rigorous.

  • @Absilicon
    @Absilicon 6 лет назад +2

    Nice, keep doing great work! 👍

  • @mohdzimam99
    @mohdzimam99 2 года назад

    Very good explanation, thanks 👍🏻

  • @ponbapanda2935
    @ponbapanda2935 4 года назад +1

    Wow i learnt so much from this video!!Amazing

  • @amaikoori
    @amaikoori 6 лет назад +10

    한국인?

  • @monke9865
    @monke9865 6 лет назад +5

    It is interesting that the derivative function is only discontinuos at one point and bounded and hence the derivatice function is riemann integrable for every interval [a, b], and in particular if you integrate the derivative over [0,x] you will just get f(x). But if you modify the function a little bit this doesnt have to happen: g(x) =x^2 sin(1/x^2) at x nonzero and g(0)=0 is differentiable function whose derivative is not riemann integrable over any [a, b] that contains 0

    • @LetsSolveMathProblems
      @LetsSolveMathProblems  6 лет назад +2

      That is a cool insight I never considered. Real analysis is full of surprises. =)

  • @Pklrs
    @Pklrs 4 года назад

    You cant have “jump” incontinuity for a derivative. It will be continious or the limit wont exist

  • @emanjoudeh859
    @emanjoudeh859 4 года назад +2

    It is exactly what I want😍

  • @ankitadas1404
    @ankitadas1404 4 года назад +1

    Show that the function x^2+sin x is continuous in all real values of x

    • @donlansdonlans3363
      @donlansdonlans3363 4 года назад +2

      Show that if f and g are continuous then so is f+g. Then show that x^2 and sinx are continuous

  • @SirTravelMuffin
    @SirTravelMuffin 2 года назад

    Thank you!

  • @rachidasavadogo8285
    @rachidasavadogo8285 Год назад

    Svp traduisez également en français

  • @harshvardhanagarwal5626
    @harshvardhanagarwal5626 4 года назад

    Sounds like Ronny Cheng

  • @tusharsrivastava6295
    @tusharsrivastava6295 3 года назад +1

    Yo

  • @a.nelprober4971
    @a.nelprober4971 3 года назад

    Easier example is modx lol

  • @UlTrAXcH
    @UlTrAXcH 4 года назад

    herroo i tanka iu for sorving dis probremu