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 $∂/∂x$?

$i$ 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.
