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For the fucntion $f:\mathbb{R}\rightarrow \mathbb{R}$ which are differentiable at $x=x_0$ imply $f'$ is continuous at $x=x_0$?

$f$ is differentiable at $x=x_0$ when $\forall \epsilon >0,\exists \delta>0, c\in \mathbb{R} \ s.t$ $$0<|x-x_0|<\delta \ \ imply \ \ |\frac{f(x)-f(x_0)}{x-x_0}-c|<\epsilon $$

in this case we let $c=f'(x_0)$

if $f'(x)$ is defined $\forall x \in\mathbb{R}$ from the above definition of derivative. can we prove $\forall \epsilon >0,\exists \delta>0,\ s.t$ $$|x-x_0|<\delta \ \ imply \ \ |f'(x)-f'(x_0)|<\epsilon $$?

I don't have any proof of above problem or counterexample, please help me!

백주상
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2 Answers2

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Differentiability does not necessarily imply the derivative is continuous. For a counterexample, see Differentiable but not continuously differentiable..

The class of differentiable functions that do have a continuous derivative are denoted $C^1$.

JKL
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$f(x)=x^{2}\sin (\frac1 x)$ for $x \neq 0$, $ f(0)=0$ defines a function which is differentiable at $0$ but its deriavtive is not continuous at $0$. In fact, $f'(x)$ does not even have a limit as $x \to 0$.

Kavi Rama Murthy
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