How Does the Chain Rule Define the Differential of a Function on a Manifold?

[Maxi1995]

We define the differential of a function f in

$$p \in M$$,

where M is a submanifold as follows

In this case we have a smooth curve ans and interval I $$\alpha: I \rightarrow M;\\ \alpha(0)= p \wedge \alpha'(0)=v$$.

How can I get that derivative at the end by using the definitions of the derivative of a function in several variables?

 

[fresh_42]

I'm not quite sure whether this will answer your question or only repeat what you already have in your book, but maybe it does (both towards the end).
See (Flows): https://www.physicsforums.com/insights/pantheon-derivatives-part-ii/
and (Pushforwards): https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/

 

[fresh_42]

For an example, you can look at Problem No. 4 here:
https://www.physicsforums.com/threads/basic-math-challenge-august-2018.952503/
with the solution here:
https://www.physicsforums.com/threads/basic-math-challenge-august-2018.952503/page-3#post-6046226
It's without an atlas, but it is what's going on between $T_p(M)$ and $T_{f(p)}N$, just define $M,N$ accordingly, i.e. in a way such that the vector fields $v,w$ of the problem becomes the tangent bundle.

 

[Maxi1995]

Thank you for your answer. So I got it right, that it is possible to interpret the differential via the chain rule as $$(f \circ\alpha)'(0)=df(\alpha(0))*\alpha'(0)=df(p)v?$$

 

[fresh_42]

You define $df(\ldots)$ in a way that the chain rule holds, so the other way around. I.e. first you get a function $g$ defined by the commutativity of
$$
\begin{equation*}
\begin{aligned}
M &\;\quad \stackrel{f}{\longrightarrow} &N\\
\downarrow{\varphi}&&\downarrow{\psi}\\
\mathbb{R}^m &\;\quad \stackrel{g}{\longrightarrow} &\mathbb{R}^n
\end{aligned}
\end{equation*}
$$
that is $g=\psi \circ f\circ \varphi^{-1}$ which you now can differentiate (with the chain rule) to define $df$. You neglected $\psi$ in your equation.