# A Differential of a function

#### 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?

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#### fresh_42

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For an example, you can look at Problem No. 4 here:
with the solution here:
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

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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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