# Why is $\left(\frac{\partial}{\partial x_i}\right)f = \left(\frac{\partial f}{\partial x_i}\right)$

i'm currently reading An Introduction to Morse Theory by Yukio Matsumoto and on p.62 it says

A vector field itself is sort of a differential operator, since it assigns to each point a "tangent vector" which is a differential operation. Let us differentiate $$f$$ with respect to the gradient vector field $$X_f$$:

$$X_f \cdot f = \left(\sum_{i=1}^m \frac{\partial f}{\partial x_i} \frac{\partial}{\partial x_i}\right)\cdot f = \sum_{i=1}^m \left( \frac{\partial f}{\partial x_i}\right)^2 \ge 0$$

I would love to understand why $$\left(\frac{\partial}{\partial x_i}\right)\cdot f = \left( \frac{\partial f}{\partial x_i}\right)$$

Could anyone help me on this? Thank you very much.

• Thank you very much @soer9606. Can you also help me understanding what "differentiating $f$ with respect to the gradient vector field" means? Because all i see is just a product of $f$ with $X_f$. – Zest Dec 15 '19 at 13:02
• @Zest I'm not completely at home with this, but I think that you can interpret it sort of like the gradient (even though the gradient and the gradient vector field is not the same). Maybe you can see some similarities in the way that they are used. Especially notice that $\nabla \cdot f = \nabla f$ where $\nabla$ is the gradient, i.e. the gradient works sort of like an action on the function. – Norse Dec 15 '19 at 13:21