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

If our function is continuous AND differentiable, does derivative have to be continuous as well?

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[LetsSolveMathProblems]

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.

[moeberry]

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

[CyrusSaiyed]

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

[lucas]

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.

[divisix]

This is the standard example in the calculus class I took

[monke]

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

[gauravjindal]

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+)

[ponbapanda]

Wow i learnt so much from this video!!Amazing

[Barman]

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 such thing is happening????

[guycohen]

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

[nadineahmed]

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

[rabeeashaheen]

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

[matteosanturi]

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

[ashudinijr]

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

[yoavshati]

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

[BB-fpce]

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

[Pklrs]

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

[ankitadas]

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

[UlTrAXcH]

herroo i tanka iu for sorving dis probremu