Tangent vectors to a plane curve Tu #11.2 This is problem 11.2 in Loring Tu's manifold text:

I don't quite understand the statement:
On the upper semicircle |U = {(a,b) ∈ S1 | b > 0}, $\bar{x}$ is a local coordinate, so that $∂/∂\bar{x}$ is defined."
So why is $∂/∂\bar{x}$ a valid derivation on $S^1$? How do we actually apply it to a smooth function defined on $U$ ( the upper semicircle)? If it is just taking the partial derivative with respect to x, then isn't $∂/∂\bar{x}$ basically just the same as $∂/∂x$?
 A: $i$ includes $S^1$ into $\mathbb R^2.$ So $i_*: T_pS^1 \to T_p \mathbb R^2.$ We have $p=(x,y)=(\bar x,\bar y)=(\bar x,\pm\sqrt{1-\bar x^2}).$ Now, there are numbers $a,b,c,d$ such that
$i_*\left(\frac{\partial }{\partial \bar x}\right)_p=a\left(\frac{\partial }{\partial x}\right)_p+ b\left(\frac{\partial }{\partial y}\right)_p\quad $ and $\quad i_*\left(\frac{\partial }{\partial \bar y}\right)_p=c\left(\frac{\partial }{\partial x}\right)_p+ d\left(\frac{\partial }{\partial y}\right)_p.$
The trick is to use the projections $r_x$ and $r_y$ to determine the constants. Namely, note that
$i_*\left(\frac{\partial }{\partial \bar x}\right)_p(r_x)=a\left(\frac{\partial }{\partial x}\right)_p(r_x)+ b\left(\frac{\partial }{\partial y}\right)_p(r_x)=a\cdot 1+b\cdot 0=a.$
Since the left-hand side of this is
$\left(\frac{\partial }{\partial \bar x}\right)_p(r_x\circ i)=\left(\frac{\partial }{\partial \bar x}\right)_p(r_\bar x)=\left(\frac{\partial }{\partial \bar x}\right)_p(\bar x)=1,$ we have $a=1.$
On the other hand,
$i_*\left(\frac{\partial }{\partial \bar x}\right)_p(r_y)=a\left(\frac{\partial }{\partial x}\right)_p(r_y)+ b\left(\frac{\partial }{\partial y}\right)_p(r_y)=a\cdot 0+b\cdot 1=b.$
Since the left-hand side of this is
$\left(\frac{\partial }{\partial \bar x}\right)_p(r_y\circ i)=\left(\frac{\partial }{\partial \bar x}\right)_p(\bar y)=\frac{\partial \bar y}{\partial \bar x},$ we have $b=\frac{\partial \bar y}{\partial \bar x}.$
Remark: a somewhat more intuitive way to do this might be simply to regard $S^1$ as the curve $c(t)=(t,\pm\sqrt{1-t^2}): 0\le t<2\pi)$ and follow the procedure outlined in sections $8.6-8.7$ in Tu's book.
A: I will use the notation of Lee's book (Introduction to Smooth Manifolds) in particular the differential of a map $F$ is denoted with $\text dF$: $(U,\varphi)$ is one chart of $S^1$ s.t. :
$$\varphi:U=\{(\bar x,\bar y)\in S^1| \bar y>0\}\to (-1,1),\;\;\varphi(\bar x,\bar y)=\bar x.$$
On the other hand $\mathbb R^2$ is the standard smooth manifold where is  well defined $\partial/\partial x,\partial/\partial y$ (or you can also view  $\mathbb R^2$ like a general manifold with the chart $(\mathbb R^2,Id)$).
The trick lies in "reading" the application on the chart  $(U,\varphi)$: $$\text di_p\left( \frac{\partial }{\partial \bar x}\bigg|_{p=(\bar x_0,\bar y_0)} \right):=\text di_p\left( (\text d\varphi)^{-1}\left(\frac{\partial }{\partial \bar x}\bigg|_{\bar x_0}\right) \right)=\text d(i\circ\varphi^{-1})_{x_0}\left(\frac{\partial }{\partial \bar x}\bigg|_{\bar x_0}\right)=*$$ Now $i\circ\varphi^{-1}:(-1,1)\to\mathbb R^2$ its differential is none other than its total derivative (https://en.wikipedia.org/wiki/Total_derivative) $\left(\begin{array}{c}\frac{\partial (i\circ\varphi^{-1})^1}{\partial\bar x} \\ \frac{(i\circ\varphi^{-1})^2}{\partial\bar x}\end{array}\right)$, but $\varphi^{-1}(\bar x)=(\bar x,\bar y)\in S^1$ (note that the second component is
$\sqrt{1-\bar x^2}$) and $i\circ\varphi^{-1}(\bar x)=(\bar x,\bar y)\in \mathbb R^2$,  so $d(i\circ\varphi^{-1})_{x_0}=\left(\begin{array}{c}1\\ \frac{\partial\bar y}{\partial \bar x}  \end{array}\right)_{x_0}$
$$*=\left(\frac{\partial}{\partial x},\frac{\partial\bar y}{\partial \bar x}\frac{\partial}{\partial y} \right).$$
I probably haven’t been totally rigorous but I try to follow the general method used in Introduction to Smooth Manifolds p.62 (He treats the general case where $i$ is a map between two general manifolds).
A: Tu usually writes charts on a smooth manifold $M$ in the form $(U,\phi) = (U,x^1,\ldots,x^n)$, where $U \subset M$ is open, $\phi : U \to \mathbb R^n$ establishes a homeomorphism between $U$ and an open subset $V \subset \mathbb  R^n$ and the $x^i: U \to \mathbb R$ are the coordinate functions of $\phi$. Sometimes he also calls $(U,\phi)$ a coordinate neighborhood although literally this only denotes the open set $U$ which is the domain of $\phi$.
On $M = \mathbb R^2$ he considers the single chart $(\mathbb R^2,x,y)$, where $x, y : \mathbb R^2 \to \mathbb R$ are the standard coordinate functions. The restrictions of $x,y$ to $S^1$ are denoted by $\bar x = i^*x = x \circ i , \bar y = i^* y = y  \circ i : S^1 \to \mathbb R$ (cf. Definition 6.1). These maps do not embed $S^1$ as open subsets of $\mathbb R$, but their restrictions to suitable subsets of $S^1$ do. In particular, the restriction of $\bar x$ to the upper open semicircle $U \subset S^1$ establishes a homeomorphism from $U$ to $(-1,1)$. A bit sloppy Tu denotes the restriction of $\bar x$ to $U$ again by $\bar x : U \to \mathbb R$. Note that this map is nothing else than the projection of $U$ to the $x$-axis.
The map $\bar x : U \to \mathbb R$ is in fact a chart on $S^1$. Tu calls it a local coordinate (which is of course true), but recall that for one-dimensional manifolds like $S^1$ charts only have a single local coordinate.
After Definition 8.1 in Chapter 3 "The Tangent Space" Tu explains that the partial derivatives $\partial/\partial x^i$ associated to a coordinate neighborhood $(U,x^1,\ldots,x^n)$ at $p$ give us derivations $\partial/\partial x^i \mid_p$ at each $p \in U$. Tu calls them coordinate vectors at $p$. In the special case $\bar x$  we get the partial derivative $\partial/\partial \bar x$ and derivations $\partial/\partial \bar x \mid_p$ at each $p \in U$. Explicitly, if $[f] \in C^\infty_p(S^1)$, then we get
$$\frac{\partial \phantom{\bar x}}{\partial \bar x} \mid_p [f] = \frac{\partial \phantom{x}}{\partial x} \mid_{\bar x(p)} (f \circ \bar x^{-1}) = \frac{d\phantom{x}}{dx} \mid_{\bar x(p)} (f \circ \bar x^{-1})  . \tag{1}$$
Here $f \circ \bar x^{-1}$ is defined on some open neighborhood of $\bar x(p)$ in $(-1,1)$. Note that the RHS of $(1)$ does not depend on the choice of the map $f$ representing the germ in $C^\infty_p(S^1)$.
In the denominator of the LHS $\bar x$ is understood as the single local coordinate on $U$, whereas it is understood as a chart on the RHS. Also note that $f \circ \bar x^{-1} : (-1,1) \to \mathbb R$ is a differentiable function in the sense of elementary calculus. We have $(f \circ \bar x^{-1})(x) = f(x,\sqrt{1-x^2})$.
Let us finally try to understand the meaning of
$$i_*\left(\frac{\partial \phantom{\bar x}}{\partial \bar x} \mid_p \right) = \left(\frac{\partial \phantom{x}}{\partial x} + \frac{\partial \bar y}{\partial \bar x}  \frac{\partial \phantom y}{\partial y} \right)_p = \frac{\partial \phantom{x}}{\partial x} \mid_p + \frac{\partial \bar y}{\partial \bar x} \mid_p \frac{\partial \phantom y}{\partial y} \mid_p . \tag{2}$$
Given $[F] \in C^\infty_p(\mathbb R^2)$, we have
$$i_*\left(\frac{\partial \phantom{\bar x}}{\partial \bar x} \mid_p \right) [F] = \frac{\partial \phantom{\bar x}}{\partial \bar x} \mid_p i^*([F]) = \frac{\partial \phantom{\bar x}}{\partial \bar x} \mid_p [F \circ i] = \frac{d\phantom{x}}{dx} \mid_{\bar x(p)} (F \circ i \circ \bar x^{-1}) . \tag {3}$$
The function $\phi = i \circ \bar x^{-1}$ is given by $\phi(x) = (x, \psi(x)))$ with $\psi(x) = \sqrt{1-x^2}$. Noting that $\phi(\bar x(p)) = p$, the chain rule gives
$$\frac{d\phantom{x}}{dx} \mid_{\bar x(p)} (F \circ i \circ \bar x^{-1}) = \frac{d\phantom{x}}{dx} \mid_{\bar x(p)} (F \circ \phi) = \frac{\partial F}{\partial x} \mid_p \frac{dx}{dx} \mid_{\bar x(p)} + \frac{\partial F}{\partial y} \mid_p \frac{d \psi(x)}{dx}  \mid_{\bar x(p)} $$ $$=   \frac{\partial F}{\partial x} \mid_p  + \frac{d \psi(x)}{dx}  \mid_{\bar x(p)} \frac{\partial F}{\partial y} \mid_p =  \left( \frac{\partial \phantom{x}}{\partial x} +  \frac{d \psi(x)}{dx}  \mid_{\bar x(p)} \frac{\partial \phantom y}{\partial \bar y}\right) \mid_p [F] . \tag {4}$$
$(3)$ and $(4)$ imply
$$i_*\left(\frac{\partial \phantom{\bar x}}{\partial \bar x} \mid_p \right) = \frac{\partial \phantom{x}}{\partial x} \mid_p +  \frac{d \psi(x)}{dx}  \mid_{\bar x(p)} \frac{\partial \phantom y}{\partial \bar y} \mid_p . \tag{5}$$
In this formula $\frac{d \psi(x)}{dx}  \mid_{\bar x(p)}$ is a real number which can be explicitly calculated.
$(5)$ looks similar as $(2)$. In fact it says the same.
The function $\bar y : S^1 \to \mathbb R$ represents the germ $[\bar y] \in C^\infty(S^1)$ and by $(1)$ we get
$$\frac{\partial \bar y}{\partial \bar x} \mid_p = \frac{\partial \phantom{\bar x}}{\partial \bar x} \mid_p [\bar y] = \frac{d\phantom{x}}{dx} \mid_{\bar x(p)} (\bar y \circ \bar x^{-1}) = \frac{d \psi(x)}{dx}  \mid_{\bar x(p)} .$$
