Differential at a point and differential (Differential Geometry) Given $f\in C^\infty(U)$, $U$ open set of $\mathbb{R}^n$, we define the differential of $f$ at $p$
$$
(df)_p:T_p\mathbb{R}^n\to\mathbb{R}\\
(df)_p(v):=v(f)
$$
and the differential of $f$
$$
df:U\to T^*U\\
p\mapsto (p,(df)_p)
$$
where $T^*U$ is the cotagent bundle.
I can understand that $(df)_p$ is the map which associates for every point $p$ any derivative $v$ to $f$. While $df$ is just obtained by gluing together all these local differentials.
Now I'm asked to compute the differential of the i-th-coordinate map $x_i:\mathbb{R}^n\to\mathbb{R}$.
Here is my (unsuccesful) reasonement.
I have to start by calculating, for every $p$, $(dx_i)_p$.
By definition $(dx_i)_p\in (T_p\mathbb{R^n})^*$ and the latter set is spanned by $\{(dx_i)_p:1\le i\le n\}$. Now, I am a bit confused. My professor defined the basis of $(T_p\mathbb{R}^n)^*$ as the dual basis of the tangent plane and the symbols $(dx_i)_p$ are just formal symbols, nothing to do with the differential. (Right?)
I would proceed by applying the differential of $x_i$ at $p$ to a general element of the tangent space $v=\sum_j v_j(\frac{\partial}{\partial x_j})_p$
$$
(dx_i)_p(v):=v_i
$$
Then how shall I go on?
 A: To remove notational confusion, let $\lambda^1,\ldots,\lambda^n$ be the dual basis of $\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^n}$. Then, your goal is to show that
$$d(x^i)=\lambda^i.$$
i.e. $d(x^i)=dx^i$, where the later is the formal symbol usually used for $\lambda^i$. Note that this justifies the notation $dx^i$.
Now, compute:
$$d(x^i)\left(\frac{\partial}{\partial x^j}\right)=\frac{\partial}{\partial x^j}x^i=\delta^i_j=\lambda^i\left(\frac{\partial}{\partial x^j}\right).$$
Hence, $d(x^i)$ agree with $\lambda^i$ on a basis, so they must be equal.
A: By whatever arguments, you yourself arrived at the correct answer: $dx_i$ computes at each point $p\in{\mathbb R}^n$ the $i$th coordinate of any tangent vector $v$ attached at $p$.
A simple way to see this is as follows: $x_i(\cdot)$ can be viewed as a real-valued function on ${\mathbb R}^n$. In order to compute the differential $dx_i(p)$ we have to look at increments
$$\Delta x_i(v):=x_i(p+v)-x_i(p)$$
and to determine an  approximation for $\Delta x_i(v)$  which is linear in $v$ when $v\to0$. Now in this special case an approximation is not needed at all, since
$$\Delta x_i(v)=v_i$$ is already linear in $v$. This implies that the differential $dx_i(p)$ is given by
$$dx_i(p).v= v_i\ ,$$
independently of the point $p$ under consideration.
