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If $f(x)$ is continuous throughout its domain, can its derivative, g(x) be non-continuous for any point?

If so, what effect does a non-continuous derivative have on the function?

dsd
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2 Answers2

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For $x\in [0,1]$, $f(0)=0$ and $f(x)=x^2\sin\frac{1}{x}$ for $x\neq 0$. Then $f(x)$ is differentiable at any point in $[0,1]$, but $f'$ is not continuous at $x=0$.

Lei Li
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Yes, consider the function $$f(x) = \begin{cases} 0 \qquad x\leq0 \\ x \qquad x>0\end{cases}.$$

It's gradient is defined everywhere except $x=0$

$$f^\prime(x) = \begin{cases}0 \qquad x<0\\1\qquad x>0\end{cases}.$$

If you want a function which is continuous but no-where differentiable Brownian Motion/Wiener Process satisfies this.

Dan
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    This function is not differentiable, as it does not have a derivative at 0. – martini Oct 14 '13 at 14:01
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    OP requires $g(x)=f'(x)$ is non-continuous for any point. – Shuchang Oct 14 '13 at 14:01
  • Ahh, so @dsd means $\forall x f^\prime(x)$ is discontinuous? – Dan Oct 14 '13 at 14:02
  • @Dan Doesn't the function have a corner point at $x = 0$ and therefore its derivative is not continuos? – dsd Oct 14 '13 at 14:04
  • @dsd: the gradient is discontinuous but the function itself is continuous as $0$. – Dan Oct 14 '13 at 14:05
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    @Dan Oops yeah. But its a corner point, so how can the derivative exist? – dsd Oct 14 '13 at 14:10
  • It doesn't that's the point. Is that not what you were after? @Downvoter, could you at least say what is wrong with my answer. – Dan Oct 14 '13 at 14:18
  • Dan, I believe the original poster wants $\exists x , f'(x)$ is discontinuous, but there's no doubt that the phrasing of the question is poor. ("for any point" should be "for some point", and it would help to move the quantifer to the beginning of the sentence.) Also, in the example you gave, $f'(x)$ is continuous at each point of its domain, and hence your $f'(x)$ is a continuous function. – Dave L. Renfro Oct 14 '13 at 20:45