Does there exist a function $f$ that is derivable at all points on its domain (derivable means first derivate is there which is continuous ) but $f''(x)$ (double derivative) does not exist at some points? Strategy: What I thought of was to think of a function which is not differentiable and then integrate it to get the desired function but I think there maybe another way of doing this without going reverse. This question I asked because I was thinking to prove by contradiction by this statement in a problem I was solving out.
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1I don't understand the question, If $f$ is derivable at all points on its domain, then there exists $f'$ at all points on domain of $f$ – ZAF Jan 29 '22 at 14:22
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f" Sir i am saying – Orion_Pax Jan 29 '22 at 14:28
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1https://math.stackexchange.com/questions/1414729/only-once-differentiable there are even functions that are differentiable everywhere once but nowhere twice. – Dan Uznanski Jan 29 '22 at 18:41
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Thanks @Dan Uznanski – Orion_Pax Jan 30 '22 at 00:42
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The key is that integrating a continuous function will give you a differentiable one: Take $f:\mathbb R\rightarrow \mathbb R$ integrable, but not differentiable (for example, $f(x)=e^{-|x|}$), then $$g(x)=\int_{-\infty}^xf(t)dt$$is differentiable with derivative $f$, but $g'$ is not differentiable at $0$ (where $f$ has a cuspid).
Alessandro
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