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Let $f,g:\Bbb R\to\Bbb R$ be real variable continuous functions and define $h:=f+g$; suppose $h$ non-constant.

If $h$ is differentiable at some $x_0\in\Bbb R$, does the same holds for both $f$ and $g$? Or there exist a couple of functions $f,g$ not differentiable at $x_0$ such that their sum does?

Maybe is trivial but I can't find counterexamples.

In what case is this true?

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    $\begingroup$ What if you take $g$ to be $-f$? $\endgroup$ – Lord Shark the Unknown Oct 4 '17 at 19:25
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    $\begingroup$ $g = -f$ where $f$ is whatever mess you want. $\endgroup$ – user296602 Oct 4 '17 at 19:25
  • $\begingroup$ Ok, that is trivial; any other counterexample? $\endgroup$ – Joe Oct 4 '17 at 19:27
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    $\begingroup$ Let $h$ be a differentiable function, $f$ be literally any non-differentiable function, and $g := h-f$. $\endgroup$ – Dustan Levenstein Oct 4 '17 at 19:28
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As others have pointed out, if you allow $f$ and $g$ to be any continuous functions, then knowing that $f+g$ is differentiable will tell you nothing about the differentiability of $f$ and $g$.

If you know that $f + g$ is differentiable and you assume that $f$ is also differentiable while making no assumptions at all on $g$, then $g$ will also be differentiable. (because $g= (f+g) - f$ and differences of diffentiable functions are differentiable.)

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If you know what a group and a subgroup are, think that the set of differentiable functions is a proper subgroup of the group of the continuous functions.

Similar examples:

If $ab>0$ are $a$ and $b$ positive?
If $a+b$ is a multiple of $3$, then both $a$ and $b$ are?
If $x,y$ are real and $x+y$ are rational, then are $x$ and $y$ rational?

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