It can be a Spivak's "Calculus" question, but I have not found anywhere so far the answer. The second derivative seems to be defined in terms of the first, but is there any way to formally prove that in general the existence of the second derivative implies the existence of the first?.
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$\begingroup$ I believe this may be a formal question. By definition, the second derivative of a function is the derivative of its derivative. If the first derivative does not exist, then how do you make sense of the second? $\endgroup$– GibbsApr 3, 2020 at 19:55
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2 Answers
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Yes, $f^{''}$ is defined to be the derivative of the derivative of $f$. so if we know $f^{''}$ exists, $f^{'}$ has to exist and be differentiable. This is true for any n>2 also.
In my opinion this is as formal as it gets, it's more a matter of definition than proof.
Assuming $f^{'}$ doesn't exist, it is obviously not differentiable (because it does not exist)
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This seems to be equivalent to asking: "If a function has a derivative, does the function exist?" Another similar question (athough maybe more obtuse): "If an element $x$ is in a set $S$, does $S$ exist?"
However, if you want a more technical reason (although there might be some circular logic going on), you could always use the fundamental theorem of calculus:
$$f'(x)-f'(a)=\int_a^xf''(t)dt$$
