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

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

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LetsSolveMathProblemsYear 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 RønbergYear 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]!

유키유노9 months ago한국인?

Quahntasy - Animating UniverseYear ago^{+3}Nice video. Loved it!

Yoav ShatiYear 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 MathsYear agoYoav 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.

LetsSolveMathProblemsYear 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.

AbeYear 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 GjorgjevskiYear 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

LetsSolveMathProblemsYear ago^{+1}That is a cool insight I never considered. Real analysis is full of surprises. =)

Benjamin WangYear 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 CenaYear agoCool try it.

Stephen BeckYear 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.

AbeYear 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).

iabervonYear agoIf 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 RahmanYear ago^{+1}Nice, keep doing great work! 👍

Matteo SanturiYear ago^{+1}I'm ok that you showed an example, but next time could you prove it by using theorems? Thanks :)

LetsSolveMathProblemsYear ago^{+8}Using a counterexample to prove that something cannot be true is completely rigorous.

Hannah WinterYear agoSome 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 kaushikYear ago^{+2}Atleast discuss STANDARD QUESTION 58 , i don't demand whole solution but atleast see and discuss it please

Siddhartha Shree kaushikYear ago^{+2}Please solve NS7UC [ standard question 58 ] I'll get you whatever money you demand

Sir Albert Einstein29 days ago@Siddhartha Shree kaushik post again

Siddhartha Shree kaushikYear ago@attyfarbuckle

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 !

attyfarbuckleYear agoYeah, 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 kaushikYear ago^{+1}@attyfarbuckle

I couldn't write whole question as LaTex format , so gave the original Source ID IN link , plus.google.com/112618997533415221768/posts/cDPg6VyZ4i3

Siddhartha Shree kaushikYear ago^{+2}@attyfarbuckle

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 !