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This question was asked in my real analysis quiz and I am unable to prove it true/ false.

State True/False -> Let f : $\mathbb{R}$-> $ \mathbb{ R} $ be a differentiable function. Then f'(x) is continuous.

I thought that f'(x) can be discontinuous as differentiability of f implies continuity of f not anything about f'(x) but I am unable to find an example. I tried sin(1/x) , cos(1/x) but it couldn't be such an example.

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    Is this question on an active quiz? – Michael Burr Jun 25 '20 at 17:55
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    This must have been covered in class, otherwise it's too hard. – zhw. Jun 25 '20 at 17:55
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    Try playing with $x^2 \sin(1/x)$ instead. Note that it is squeezed between two functions $-x^2$ and $x^2$, which are differentiable at $0$. – user803264 Jun 25 '20 at 17:57
  • OK here is the general rule of thumb in real-analysis. If you have not proved $X$ or seen $X$ as a standard theorem then most likely $X$ is false. Don't rely on typical examples. – Paramanand Singh Jun 25 '20 at 17:58
  • @Michael Burr why are you thinking that? –  Jun 25 '20 at 18:10
  • @Parmanand Singh I was asked to provide a counterexample if it's false so can you provide any? –  Jun 25 '20 at 18:24
  • @user Didn't you want a differentiable function with non-continuous derivative? What's wrong with user803264's example? – Ned Jun 25 '20 at 18:27
  • @user Your question stated that this was from a quiz. But you didn't state whether this was from a quiz that you had taken (and turned in) or whether it was from a quiz that you were currently taking. Getting help on active assignments quizzes/exams/competitions is typically frowned upon here. – Michael Burr Jun 25 '20 at 22:58

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It is false, you can use this example: $$\ f(x) = \begin{cases} x^2 \sin(\frac {1}{x}) & \mbox{ if } x \ne 0 \\ 0 & \mbox{ if } x = 0 \end{cases}$$ so as you can see $f'(0)=0$ but $${\displaystyle \lim _{x\to 0}f'(x)=\lim _{x\to 0}\left[2x\sin(\frac {1}{x})-\cos(\frac {1}{x})\right]\neq 0 \ \ (the \ limit \ doesn't \ exist)}$$ so $f'$ is not continuous at $0$.

The student
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