TITLE: Use of partial derivatives as basis vector
QUESTION [22 upvotes]: I am trying to understand use of partial derivatives as basis functions from differential geometry

In tangent space $\mathbb{R^n}$ at point $p$, the basis vectors $e_1, e_2,...,e_n$ can be written as $\frac {\partial}{\partial x^1} \bigg|_p,\frac {\partial}{\partial x^2} \bigg|_p,...,\frac {\partial}{\partial x^n} \bigg|_p$

Let's say in 2 dimensional Euclidean space, a function $f : \mathbb {R^2}\rightarrow \mathbb {R^2}$ is
$x^2 + y^2=4$ , a circle with radius 2. 
Tangent at point $p$ (2,0) will be $0e_1 + e_2$. 
If I say $f =x^2 + y^2-4 =0$, 
$\frac {\partial f}{\partial x} \bigg|_{p=(2,0)} = 4 \quad$ and   $\quad \frac {\partial f}{\partial y} \bigg|_{p=(2,0)} = 4$
This does not make sense of the partial derivatives as basis vectors. Any comments?

REPLY [4 votes]: The space of the OP's circle is not a tangent space, that's why the simple example does not make any sense. One must stay inside the circle, so to speak. So let's start over again and define that circle with radius $2$ as:
$$
\begin{cases} x = 2\cos(\phi) \\ y = 2\sin(\phi) \end{cases} \quad \Longrightarrow \quad x^2+y^2=4
$$
There are no partial derivatives because the manifold is one-dimensional.
And there is only one basis (tangent) vector:
$$
\begin{bmatrix} dx/d\phi \\ dy/d\phi \end{bmatrix} = \begin{bmatrix} -2\sin(\phi) \\ 2\cos(\phi) \end{bmatrix}
$$
Normed to a unit vector $\,\vec{t}\,$ if we divide by $2$ :
$$
\vec{t} = \begin{bmatrix} dx/d\phi \\ dy/d\phi \end{bmatrix} / 2 =
\begin{bmatrix} -\sin(\phi) \\ \cos(\phi) \end{bmatrix} = -\sin(\phi)\,\vec{e_1} +  \cos(\phi)\,\vec{e_2}
$$
Now specify for $\,\phi=0\,$ and we're done, for this example at least.