A Differential of a function

14
0
We define the differential of a function f in

$$p \in M$$,

where M is a submanifold as follows
Untitled01.jpg

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?
 

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fresh_42

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fresh_42

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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.
 
14
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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

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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.
 

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