TITLE: How can the derivative of $f:M\to \mathbb R$ where $M$ is a differentiable manifold can be well defined since it's not unique?
QUESTION [7 upvotes]: I have a doubt about the fact that a derivative of $f:M\to \mathbb R$ of a $\mathcal C^1$ manifold is well defined... Indeed, let $a\in M$ and $(U,\varphi )$ a chart from a $\mathbb C^1$ atlas s.t. $a\in A$. Then, $$f'(a)=\left.\frac{d}{dx}\right|_{x=\varphi (a)}f(\varphi ^{-1}(x)).$$
But if $(\psi,V)$ is an other chart s.t. $a\in V$, then, $$f'(a)=\left.\frac{d}{dx}\right|_{x=\psi(a)}f(\psi^{-1}(x)).$$
The thing is that I don't see any reason to have $$\left.\frac{d}{dx}\right|_{x=\varphi (a)}f(\varphi ^{-1}(x))=\left.\frac{d}{dx}\right|_{x=\psi(a)}f(\psi^{-1}(x)).$$

REPLY [5 votes]: While the derivative of $f$ in a coordinate system is map-dependent, there's a way to define it without using any map or coordinates.
In such a definition, the derivative of a function $f: M\rightarrow \mathbb R$ at point $p\in M$ is a covector  $(df)_p \in {T_p}^*M$ such that for any curve $\gamma : \mathbb R \rightarrow M$, such that $\gamma(0) = p$ the following holds:
$$ \left.\frac{d}{dt}\right|_{t=0}f(\gamma(t)) = \langle (df)_p, \gamma'(0) \rangle $$
where $\gamma'(0) \in T_pM$ is the vector tangent to curve $\gamma$ at point $p$.
You can check that this defines $(df)_p$ in a unique way, and if you choose any coordinate system, then the coefficients of $(df)_p$ in the induced basis in ${T_p}^*M$ are equal to the partial derivatives of $f$ in respect to chosen coordinates.