# Differentiable function

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• November 13th 2009, 05:47 AM
hjortur
Differentiable function
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.
Then:
$f(x)=(g\circ f')$
$(g^{-1}\circ f)(x)=g^{-1}\circ(g\circ f')(x)=f'(x)$

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.