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

  • Published on Sep 16, 2018
  • If our function is continuous AND differentiable, does derivative have to be continuous as well?
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Comments • 28

  • LetsSolveMathProblems
    LetsSolveMathProblems  9 months ago +3

    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.

    • Lauge Rønberg
      Lauge Rønberg 9 months ago +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]!

  • 유키유노
    유키유노 6 months ago


  • Quahntasy - Animating Universe

    Nice video. Loved it!

  • Yoav Shati
    Yoav Shati 9 months ago +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

    • RGP Maths
      RGP Maths 9 months ago

      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.

    • LetsSolveMathProblems
      LetsSolveMathProblems  9 months ago +1

      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.

    • Abe
      Abe 9 months ago +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

  • Martin Gjorgjevski
    Martin Gjorgjevski 9 months ago +3

    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  9 months ago +1

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

  • Benjamin Wang
    Benjamin Wang 9 months ago +5

    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.

    • John Cena
      John Cena 9 months ago

      Cool try it.

    • Stephen Beck
      Stephen Beck 9 months ago +2

      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.

    • Abe
      Abe 9 months ago +1

      I guess it's because that makes it so every sine/cosine has an x factor. If you change x^3sin(1/x) to x^3sin(1/x^2), then f*g' (using the product rule) will be x^3*cos(1/x^2)*(-2/x^3)=-2*cos(1/x^2). Without being multiplied by a function that approaches 0, there's no way that cosine is gonna behave nicely. So, if you have f(x)=x^n*sin(1/x^m): if n,m are integers and n>=m+2, then f'(x) will be continuous and differentiable...I think (you also need to do some special stuff at x=0 like he did in the video, probably, but I think what I said is valid).

    • iabervon
      iabervon 9 months ago

      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.

  • Abidur Rahman
    Abidur Rahman 9 months ago +1

    Nice, keep doing great work! 👍

  • Matteo Santuri
    Matteo Santuri 9 months ago +1

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

    • LetsSolveMathProblems
      LetsSolveMathProblems  9 months ago +8

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

    • Hannah Winter
      Hannah Winter 9 months ago

      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.

  • Siddhartha Shree kaushik
    Siddhartha Shree kaushik 9 months ago +2

    Atleast discuss STANDARD QUESTION 58 , i don't demand whole solution but atleast see and discuss it please

  • Siddhartha Shree kaushik
    Siddhartha Shree kaushik 9 months ago +2

    Please solve NS7UC [ standard question 58 ] I'll get you whatever money you demand

    • Siddhartha Shree kaushik
      Siddhartha Shree kaushik 9 months ago

      You've no idea what actually this question is.
      This question is next generation advanced LEVEL MATHEMATICS
      Beyond the reach of quantum computation values and super computers ,
      I want a perfect discussion on it !

    • attyfarbuckle
      attyfarbuckle 9 months ago

      Yeah, maybe LSMP will take this on. I posted that link because I wanted something that could be copy/pasted somewhere else like a graphing program or solver or search engine without worrying about missing a minus sign or bracket.
      Note though if there is a general solution, it is at least as complicated as the case when a = 2.
      And that has minimal polynomial: 113 - 776 x^2 + 1584 x^4 - 1152 x^6 + 256 x^8
      So this might take more than a 10 minute video to explain.

    • Siddhartha Shree kaushik
      Siddhartha Shree kaushik 9 months ago +1

      I couldn't write whole question as LaTex format , so gave the original Source ID IN link ,

    • Siddhartha Shree kaushik
      Siddhartha Shree kaushik 9 months ago +2

      Thanks a lot for your efforts
      But keep in mind that x ∈ [ -a , +a ]
      moreover a=x=0 is considered invalid result and believe me , because we demanded x in terms of a , and this question is really a monster or devil in this universe
      I'm praying that somehow this USclip channel helps me out !

    • attyfarbuckle
      attyfarbuckle 9 months ago +1

      Siddhartha, what have you tried so far? I have no general solution, but if you put a = 2 there is this closed form solution that might hint at a full answer.
      x = sqrt(9/8 + 1/sqrt(2) - 1/8 sqrt(13 + 8 sqrt(2))) solves this.
      Good question by the way, but if you want help with something like this it is easier if you can post the question in some readable way.