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Assume that $f(x)$ and $g(x)$ are both continuous real functions, and that, for $x \ne 0$, $f'(x) = g(x)$.

Is then $f(x)$ differentiable at $x = 0$, with $f'(0) = g(0)$?

J. W. Tanner
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Loic
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1 Answers1

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Yes. By the MVT, we have $$\dfrac{f(0+h)-f(0)}{h} = f'(c(h))=g(c(h))$$ where $0<c(h)<h.$ Taking the limit as $h \to 0$ of both sides yields that $f$ is differentiable with $f'(0)=g(0).$

The reason why we are able to conclude that $\lim_{h \to 0} g(c(h))= g(0)$ is because $g$ is continuous, so $\lim_{h \to 0} g(c(h))= g(\lim_{h \to 0} c(h))=g(0).$

Ovi
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  • I think it's worth elaborating on the last step when taking the limit $h \to 0$, because there, we need to make use of continuity of $g$ at $0$ (and that of course a similar argument works if $h<0$). – peek-a-boo Sep 22 '19 at 18:52
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    I think it would be perhaps more convincing to look at the limit $|{f(h)-f(0) \over h} - g(0)| $. The existence of a function $c$ would need to be established and is unnecessary. – copper.hat Sep 22 '19 at 18:54
  • @peek-a-boo Done – Ovi Sep 22 '19 at 18:54
  • It is certainly true that for any $h$ there is some $\eta_h \in (0,h)$ such that the quotient above equals $g(\eta_h)$, but there are some logical steps needed to demonstrate that we can define a function $c(h) = g(\eta_h)$. – copper.hat Sep 22 '19 at 18:57
  • @copper.hat Hmm what are the extra logical steps? – Ovi Sep 22 '19 at 18:58
  • Well, not quite axiom of choice, but more argument that is necessary, see https://math.stackexchange.com/a/1051775/27978 for example. Just stick with the existence of $\eta_h$ ofr simplicity. – copper.hat Sep 22 '19 at 19:02
  • @copper.hat Thanks for the link! – Ovi Sep 22 '19 at 19:03