Differential of a function

  • #1
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?
 

Attachments

Answers and Replies

  • #3
13,457
10,516
  • #4
14
0
Thank you for your answer. So I got it right, that it is possible to interpret the differential via the chain rule as $$(f \circ\alpha)'(0)=df(\alpha(0))*\alpha'(0)=df(p)v?$$
 
  • #5
13,457
10,516
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.
 
Top