Prove that $(dx^i)_p\Bigg( \frac{\partial}{\partial x^j}|_p \Bigg )=\delta ^i_j$ In Tu's book, An Introduction to Manifolds (Second edition), at page 35 states that by definition:
$$(dx^i)_p\Bigg( \frac{\partial}{\partial x^j}|_p \Bigg )=\frac{\partial}{\partial x^j}|_p x^i$$
I don't understand how is that by definition.
If I apply the definition, I get:
$$(dx^i)_p\Bigg( \frac{\partial}{\partial x^j}|_p \Bigg )=\sum_{j=1}^n \frac{\partial}{\partial x^j }|_p \frac{\partial x^i}{\partial x^j}|_p$$
why is this true I don't follow  $$\sum_{j=1}^n \frac{\partial}{\partial x^j }|_p \frac{\partial x^i}{\partial x^j}|_p=\frac{\partial}{\partial x^j}|_p x^i$$
Can someone explain ?

Exterior derivative of a function is defined through the directional derivative:
$df(X)=X(f)$, where $X$ is an arbitrary tangent vector. We apply that formula on a coordinate function $f=x^i$ and along a tangent to the j-coordinate curve, $X=\frac{\partial}{\partial x^j}$, all happening at a point $p$.
In order to calculate the directional derivative $\frac{\partial}{\partial x^j}(x^i)$ on a manifold we're going to need a curve passing through the point $p$ at $t=t_0$ with the tangent velocity $\frac{\partial}{\partial x^j}$, the aforementioned j-coordinate curve is precisely that at each point of our chart. Let me denote the j-coordinate curve function as $x^{-j}\equiv({x^j})^{-1}$, it's the inverse to the j-coordinate function: $x^j\circ x^{-j}\equiv id$, but all other coordinates functions are constant on this curve. So we have,
$$\frac{\partial}{\partial x^j}(x^i)|_p=\lim_{t\to 0}\frac{[x^i\circ x^{-j}]_{t_0+t}-[x^i\circ x^{-j}]_{t_0}}{t}=\delta ^i_j$$
