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Let $M^n$ be a smooth manifold and $(U,\phi)$ be a chart with local coordinates $x^1,...,x^n$, that is $\phi(u)=(x^1(u),...,x^n(u))$, with $x^i:U\to\mathbb{R}$. For $f\in C^\infty(M)$ we define $$\tfrac{\partial}{\partial x^i}f:U\to \mathbb{R},\quad p\mapsto\tfrac{\partial}{\partial x^i} f(p)=\partial_i(f\circ\phi^{-1})(\phi(p)), $$ where the partial derivative is w.r.t. the $i$-th coordinate is just partial derivative in $\mathbb{R}^n$. Now if we take a second chart $(V,\psi)$ with local coordinate functions $y^1,...,y^n$, why is it true that for $p\in U\cap V$ one has $$\tfrac{\partial}{\partial x^i}|_p=\tfrac{\partial}{\partial x^i}|_p(y^j)\tfrac{\partial}{\partial y^j}|_p=\sum_{j=1}^n \tfrac{\partial}{\partial x^i}|_p(y^j)\tfrac{\partial}{\partial y^j}|_p. $$ (In the identity the Einstein notation for sums was used). I think, I understand why it makes sense to define the partial derivative like this and for the case of $M^n=\mathbb{R}^n$, we can just take $x^i$ and $y^i$ to be the standard coordinates, in which case the identity would be trivial. But how to argue for general manifolds?

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  • $\begingroup$ Use the fact that $\frac{\partial}{\partial y^1}|_p, \ldots, \frac{\partial}{\partial y^n}|_p$ is a basis for the vector space $T_p M$. $\endgroup$
    – Uncool
    Apr 27 at 7:39
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Since $\frac{\partial}{\partial y^1}|_p, \ldots, \frac{\partial}{\partial y^n}|_p $ is a basis for the vector space $T_pM$, we can express each vector $\frac{\partial}{\partial x^i}|_p \in T_p M $ as a linear combination $$ \dfrac{\partial}{\partial x^i}\bigg|_p = a_1 \dfrac{\partial }{\partial y^1}\bigg|_p + \cdots + a_n \dfrac{\partial }{\partial y^n}\bigg|_p ,$$ for some $a_1, \ldots, a_n \in \mathbb{R}$. Now to find out the constants $a_j$, we apply both sides of the above identity on the functions $y^j$: $$\dfrac{\partial }{\partial x^i}\bigg|_p y^j= a_1 \dfrac{\partial }{\partial y^1}\bigg|_p y^j + \cdots + a_n \dfrac{\partial }{\partial y^n}\bigg|_p y^j.$$ Because $\frac{\partial }{\partial y^i}|_p y^j= \delta^{i}_j $, we get that $$\dfrac{\partial }{\partial x^i}\bigg|_p y^j= a_j.$$ Therefore substituting these values of $a_j$ on the first identity we get $$\dfrac{\partial}{\partial x^i}\bigg|_p = \dfrac{\partial y^1}{\partial x^i}(p)\dfrac{\partial}{\partial y^1}\bigg|_p + \cdots + \dfrac{\partial y^n}{\partial x^i}(p)\dfrac{\partial}{\partial y^n}\bigg|_p,$$ as required.

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