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So the theorem is

$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$

The usual proof on Wikipedia or here seems to come from defining a $h(x)$ and doing some magic with it.

But with the mean value theorem $f'(c) = \frac{f(b) - f(a)}{b-a}$, can't this simply be proved by bringing this on a form like $b-a=\frac{f(b)-f(a)}{f'(c)}=\frac{g(b)-g(a)}{g'(c)}$ and rearranging this to the form at the top? I'm probably missing something crucial reason.

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$c$ is not the same here. You have $f'(c_1)=\frac{f(b)-f(a)}{b-a}$ for some $c_1\in\mathbb R$ and you have $g'(c_2)=\frac{g(b)-g(a)}{b-a}$ for some $c_2\in\mathbb R$

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