Question about $\partial_i \left(f \circ x^{-1}\right)(x(p))$

Question

In this video(at about 55:00), the lecturer wrote $\partial_i \left(f \circ x^{-1}\right)(x(p))$ as $\left (\dfrac{\partial f}{\partial x^{i}} \right)_p$. I am very confused about this notation. The lecturer says its not a partial derivative, but how is it useful to just denote like this? $x$ is a function in $\partial_i \left(f \circ x^{-1}\right)(x(p))$ but how can it be like a coordinate in $\left (\dfrac{\partial f}{\partial x^{i}} \right)_p$? Does it have something to do with partial derivatives? I would appreciate it if there's someone could clarify this for me.

Answer 1

It is useful at least in two aspects.

When the smooth manifold is $\mathbb{R}^n$ and if one considers the identity map as the coordinate map, then $\partial_i(f\circ x^{-1})(x(p))$ coincide with the partial derivative in the classical sense.

For general manifolds, this notation reminds you of the name of the coordinate map and the direction of the partial derivative for $(f\circ x^{-1})$.

By definition of the video, $$ \left(\frac{\partial f}{\partial x^i}\right)_p:=\partial_i(f\circ x^{-1})(x(p)) $$

What the lecturer means is that it is not a partial derivative for the function $f$, because it does not make sense to talk about "partial derivatives" for functions defined on a general manifold $M$; for instance, you cannot "add" two elements on $M$. But it is a partial derivative for the function $f\circ x^{-1}$, which is a function on the Euclidean space.