0

Prove or disprove:

There exists a function $f : [−1, 1] \to \mathbb{R}$ such that $f$ is differentiable for every $x \in [−1, 1]$, but $f'$ is not continuous on $[−1, 1]$.

J. W. Tanner
  • 60,406
Aviv Barel
  • 45
  • 5
  • This is a duplicate of THIS. – Mark Viola Jan 22 '20 at 19:28
  • 1
    @MarkViola: that question was asking for examples where the derivative is discontinuous at more than one point – J. W. Tanner Jan 22 '20 at 19:30
  • @J.W.Tanner No, it does not ask anything about multiple points of discontinuity. Why on earth do you infer that? – Mark Viola Jan 22 '20 at 19:34
  • 1
    @MarkViola: because the question you linked states as follows: "Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point", but I concede that any answer to that question would surely answer this question – J. W. Tanner Jan 22 '20 at 19:47

2 Answers2

2

hint

Take $$f(x)=x^2\sin(\frac{1}{x^2})$$ for $x\ne 0$ and $f(0)=0$.

$f$ is continuous at $[-1,1]$.

$f$ is differentiable at $[-1,1]$

But

For $x\ne 0$

$$f'(x)=2x\sin(\frac{1}{x^2})-\frac{2}{x}\cos(\frac{1}{x^2})$$

which has no limit at $x=0$.

hamam_Abdallah
  • 62,951
0

One such example is $$f(x) = \arcsin x$$

since $$f'(x) = {1\over \sqrt{1-x^2}}$$

nonuser
  • 90,026
  • 1
    This example is not quite correct, since $f$ is not differentiable on the closed interval. I think the question is about finding a discontinuity of $f'$ inside the interval (at least that is more interesting). – GReyes Jan 22 '20 at 19:25
  • how do i prove arcsin is differentibale – Aviv Barel Jan 22 '20 at 19:55