TITLE: Directional derivative of smooth real values functions wrt. smooth curves
QUESTION [1 upvotes]: Let $M$ be a $m$-dimensional smooth manifold, $p \in M$ a point and consider smooth curves $\alpha: (-\epsilon,\epsilon) \to M$, $\beta: (-\tilde{\epsilon},\tilde{\epsilon}) \to M$ centered at $p$, meaning $\alpha(0)=\beta(0)=p$.
We define the following equivalence relation on these curves:
$\alpha \sim \beta :\Longleftrightarrow $ for every chart $\varphi: U \to U'$ with $p \in U$ we have $(\varphi \circ \alpha)'(0) = (\varphi \circ \beta)'(0)$.
We call an equivalence class a tangent vector.
Let $\alpha$ be such a curve  through $p \in M$ and $f: V \to \mathbb{R}$ a smooth real valued function defined on an open neighborhood $V$ of $p$.
Then we call $(f \circ \alpha)'(0)$ the (directional) derivative of $f$ in direction $\alpha$. 
My notes say that this derivative only depends on the equivalence class of $\alpha$, so this must mean that if $\alpha \sim \beta$ we have $(f \circ \alpha)'(0) = (f \circ \beta)'(0)$. But I do not understand why this holds.
I know that if $\alpha \sim \beta$ we have $(\varphi \circ \alpha)'(0) = (\varphi \circ \beta)'(0)$ for any chart $\varphi: U \to U'$ defined on an open neighborhood $U$ in $p$. But I do not know how to make the transition from $\varphi$, which maps to an open subset $U' \subset \mathbb{R}^m$, to $f$ which only maps to $\mathbb{R}$.
Could you please explain this problem to me? Thank you!

REPLY [3 votes]: Just note that $(f \circ \alpha) = (f \circ \varphi^{-1} \circ \varphi \circ \alpha)$ and $\alpha(0)=\beta(0)=p$ and $(\varphi \circ \alpha)'(0) = (\varphi \circ \beta)'(0)$.  So by the chain rule
\begin{align}
(f \circ \alpha)'(0) &= (f \circ \varphi^{-1} \circ \varphi \circ \alpha)'(0) \\
&= D(f \circ \varphi^{-1})_{\varphi(p)} \circ (\varphi \circ \alpha)'(0)\\
&= D(f \circ \varphi^{-1})_{\varphi(p)} \circ (\varphi \circ \beta)'(0)\\
&= (f \circ \varphi^{-1} \circ \varphi \circ \beta)'(0) \\
&= (f \circ \beta)'(0).
\end{align}
In fact, this definition of directional derivatives can be extend for all global function $ f \in C^{\infty}(M)$ using bump function argument. See Jeff. Lee or here.
{"set_name": "stack_exchange", "score": 1, "question_id": 2569816}