A differentiable bijection $f$ from $I$ (real interval different from an empty one and a point) to $J$.
$f'$ is the derivative of $f$ and a bijection from $I'$ to $J'$.
$g$ is the inverse of $f$ from $J$ to $I$.
And $g'$ has to be the derivative of $g$ and the inverse of $f'$ (from $J'$ to $I'$).
Is there a function $f$ that verifies this condition ?
My conjecture is that there is no such a function.
Thanks for help.
On
I will assume $f$ is twice differentiable (EDIT: and increasing). As $f'$ is a bijection, either $f''>0$ or $f''<0$ everywhere. So either $f$ is convex or concave. This means its inverse $g$ is concave or convex, respectively, and $g''$ has the opposite sign to $f''$. However, as $f'$ and $g'$ are inverse to each other, $g''(t)=\frac1{f''(g'(t))}$ so $f''$ and $g''$ have the same sign everywhere. That is a contradiction.
EDIT: If the function is decreasing, there is no problem with the signs, as pointed out in the comments below.
NOT a complete answer, but just some ramblings to clean up the problem a bit.
First, $f'$ is defined on the same interval as $f$, so your $I'$ is just $I$; I'm going to use $K$ to denote $f'(I)$, rather than using $J'$.
You want $$ g'(f'(u)) = u $$ for all $u \in I$. But because $g = f^{-1}$, we know (inverse function theorem) that $$ g'(b) = \frac{1}{f'(f^{-1}(b))} $$ for any $b \in K$. Applying this to $b = f'(u)$, and substituting, we get $$ g'(f'(u)) = \frac{1}{f'(f^{-1}(f'(u)))} $$ From the first equation, you'd like this to equal $u$, for all $u \in I$, i.e., you want to find a function $f$ on some interval $I$ with the property that $$ \frac{1}{f'(f^{-1}(f'(u)))} = u, $$ or $$ \frac{1}{u} = f'(f^{-1}(f'(u))) $$
Now in this equation, you're applying $f^{-1}$ to $f'(u)$, so that means that $K \subset J$. So the problem becomes:
Are there intervals $I$ and $J$ and a function $f: I \to J$, with
Personally, I suspect not, but I don't have any sort of proof.