Does the everywhere differentiability of $f$ imply it is absolutely continuous on a compact interval? Suppose $f$ is differentiable everywhere on $[0,1]$. Must $f$ be absolutely continuous on $[0,1]$? I know this is true if $f'$ is integrable but I'm not sure in this more general case.
No.  Consider $f(x) = x^2 \sin(1/x^4)$ on $[-1,1]$.  Note that for $\beta_n = (\pi n)^{-1/4}$ and $\alpha_n = (\pi (n+1/2))^{-1/4}$ we have $\beta_n - \alpha_n = \Theta( n^{-5/4})$ while $|f(\beta_n) - f(\alpha_n)| = \Theta(n^{-1/2})$.
$ f ( x ) = \frac 1 { x + \epsilon } $, $ \epsilon $ tiny positive value

is differentiable every where in $[0,1]$ but not uniformly continuous which implies it is not absolutely continuous either.