$f( \phi^{-1}(x_0 +h)) = f(\phi^{-1}(x_0+h))+h \alpha+ O(h^2)$ - value of $\alpha$

Question

Consider a function (continuous) $f : M \to \mathbb{R}$ with $M$ a $1$-dimensional manifold, and suppose some (smooth) chart $\phi : T \to \mathbb{R}$ having an (smooth) inverse on some open $(x_0- \epsilon, x_0 + \epsilon) \subset \mathbb{R}$, and that when (1) $h \to 0$ : $f( \phi^{-1}(x_0 +h)) = f(\phi^{-1}(x_0))+h \alpha+ O(h^2)$. Is there anyone could be able to explain how to obtain (1) and what could be the value of $\alpha$?

Answer 1

This is a Taylor expansion of the function in local coordinates. You should prove that in fact $f$ is $C^1$, not just continuous, and $\alpha$ should be the derivative.