Let f and g be differentiable functions for all x\in\mathbb{R}
Now if f(x)=(g\circ f')(x), can I conclude that because f is
differentiable, thus (g\circ f') is differenitable. And further
that if (g\circ f') is differentiable then: (g\circ
f')'=g(f'(x))\cdot f''(x) and thus f is twice differentiable?
If not, then how about if g is bijective and thus has an inverse
g^{-1}, and the inverse is differentiable. f(x)=(g\circ f')
(g^{-1}\circ f)(x)=g^{-1}\circ(g\circ f')(x)=f'(x)  And now
g^{-1}\circ f is a combination of 2 differentiable functions,
thus (g^{-1}\circ f)'=\left(f'(x)\right)'=f''(x) then f is twice
differentiable?  This is not homework, I am just thinking.