With $X, Y$ vector fields and $f$ a smooth function, show that $X(gY) = (Xg)Y + gXY$

Question

I know that $X = \sum_i a_i \frac \partial\partial_{x_i}$ and $Y=\sum_j b_j \frac \partial\partial_{x_j}$ but I'm not sure how to proceed. The only approach I can think of is something to do with the chain rule.

Answer 1

Hint:

$$XY=\sum_i a_i \frac{\partial}{\partial{x_i}}\left(\sum_j b_j\frac {\partial}{\partial x_j}\right),$$

$$=\sum_{ij} a_i \frac{\partial}{\partial{x_i}}\left(b_j \frac {\partial}{\partial{x_j}}\right),$$

$$=\sum_{ij} a_i \left(\frac{\partial b_j}{\partial{x_i}}\frac{ \partial}{\partial{x_j}} +b_j\frac{\partial}{\partial{x_i}}\frac{\partial}{\partial{x_j}}\right),$$

$$=\sum_{ij} a_i \left(\frac{\partial b_j}{\partial{x_i}}\frac{\partial}{\partial{x_j}}+b_j\frac{\partial}{\partial{x_i}}\frac{ \partial}{\partial{x_j}}\right),$$

$$=\sum_{ij} a_i\frac{\partial b_j}{ \partial{x_i}}\frac{ \partial}{\partial{x_j}}+\sum_{ij} a_ib_j\frac{\partial}{\partial{x_i}}\frac{ \partial}{\partial{x_j}}.$$