Let $\displaystyle f$ and $\displaystyle g$ be differentiable functions for all $\displaystyle x\in\mathbb{R}$

Now if $\displaystyle f(x)=(g\circ f')(x)$, can I conclude that because $\displaystyle f$ is differentiable,
thus $\displaystyle (g\circ f')$ is differenitable.
And further that if $\displaystyle (g\circ f')$ is differentiable then:
$\displaystyle (g\circ f')'=g(f'(x))\cdot f''(x)$ and thus $\displaystyle f$ is twice differentiable?

If not, then how about if $\displaystyle g$ is bijective and thus has an inverse $\displaystyle g^{-1}$, and the inverse is differentiable.
Then:
$\displaystyle f(x)=(g\circ f')$
$\displaystyle (g^{-1}\circ f)(x)=g^{-1}\circ(g\circ f')(x)=f'(x)$

And now $\displaystyle g^{-1}\circ f$ is a combination of 2 differentiable functions, thus
$\displaystyle (g^{-1}\circ f)'=\left(f'(x)\right)'=f''(x)$
then $\displaystyle f$ is twice differentiable?

This is not homework, I am just thinking.