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If a function $f$ is defined and differentiable in an open interval A, then it's also continuous in A. Since it's differentiable, for a random real number a that belongs to A:

$f'(a) = \lim_{x\to a}{f(x) - f(a)\over x - a} $

Since $f$ is continuous, $\lim_{x\to a}f(x) = f(a)$ so the limit $\lim_{x\to a}{f(x) - f(a)\over x - a}$ is ${0\over 0}$, so we can use L'Hospital's rule and say that $f'(a) = \lim_{x\to a}{f(x) - f(a)\over x - a} = \lim_{x\to a}{f'(x)}$

(pretty much the same thing if it's a closed interval)

So that means that $f'$ is continuous, which is wrong since it's just a random function and it's possible to make a derivative of a function non continuous. So where's the mistake?

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    L'Hospital's rule says that (besides the other premises) **if** $\lim\limits_{x\to a} \dfrac{f'(x)}{g'(x)}$ exists, then $\lim\limits_{x\to a} \frac{f(x)}{g(x)}$ exists and equals the former. It doesn't say that the former limit exists if the latter exists. – 2017-02-05
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    Oh, I see. Thanks! – 2017-02-05

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You can not use l'Hospital here because you need to know about the existence of $$\lim_{x\to a} f'(x)$$ first. You are messing the order of implication of l'Hospital. To show that a differentiable function is continuous have a look at $$f(x)-f(a)= \frac{f(x) -f(a)}{x-a} \cdot (x-a)$$ and use the known limit results.

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    Is there a mistake in that expression? – 2017-02-05
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    I can't see one :P Corrected it, thank you! :) – 2017-02-05