Source: Prof. Meinrenken's notes on Differential Geometry, page No 72:
Theorem $4.1$. let $(U,\varphi)$ be a coordinate chart around $p$.A linear map $v :C^{\infty}(M) \to \mathbb{R}$ is in $T_pM$ if and only if it has the form $$v(f)=\sum_{i=1}^{m} a^i\frac{\partial(f \circ \varphi^{-1})}{\partial u^i}|_{u=\varphi(p)}$$ for some $a=(a^1,…..,a^m)\in \mathbb{R}^m$
Proof: Given a linear map v of this form.Let $\bar{\gamma}: \mathbb{R} \to \varphi(U)$ be a curve with $\bar\gamma(t)=\varphi(p) +ta $ for $|t|$ suffiently small .Let $\gamma = \varphi^{-1} \circ \bar\gamma$.Then
$$\frac{d}{dt}|_{t=0} f(\gamma(t))=\frac{d}{dt}|_{t=0}(f\circ \varphi^{-1})(\varphi(p) +ta)=\sum_{i=1}^{m} a^i\frac{\partial(f \circ \varphi^{-1})}{\partial u^i}|_{u=\varphi(p)} $$
by chain rule
My confusion:How to use chain rule in this equation
why $$\frac{d}{dt}|_{t=0}(f\circ \varphi^{-1})(\varphi(p) +ta)=\sum_{i=1}^{m} a^i\frac{\partial(f \circ \varphi^{-1})}{\partial u^i}|_{u=\varphi(p)} ?$$
I know that chain rule mean
$$\frac{\partial (f \circ F)}{\partial x^i} =\sum_{j=1}^m\frac{\partial f}{\partial y^j}\frac{\partial F^j}{\partial x^i}.$$
where $f:V\to\Bbb R$ on an open set $V\subseteq\Bbb R^m$, with its partial derivative with respect to the $j$-th coordinate denoted $\frac{\partial f}{\partial y^j}$.
and $F:U\to V$ on an open set $U\subseteq\Bbb R^n$, with its $j$-th component function's partial derivative with respect to the $i$-th coordinate denoted $\frac{\partial F^j}{\partial x^i}$.
My attempt: $$\frac{d}{dt}|_{t=0}(f\circ \varphi^{-1})(\varphi(p) +ta)=\frac{d}{dt}|_{t=0}((f\circ \varphi^{-1})(\varphi(p)) + (f\circ \varphi^{-1})ta)=\frac{d}{dt}|_{t=0}((f(p)) + (f\circ \varphi^{-1})ta) $$
After that Im not able to proceed further
Best Answer
You wrote
This step is assuming e.g. linearity of $f\circ \varphi^{-1}$ which is not known.
You just need to apply your stated version of chain rule for differentiating $f\circ F$, replacing $f$ with $f\circ \varphi^{-1}$, and putting $F(t)=\gamma(t)=\phi(p)+ta$.