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

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.

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How do you understand $XY$? –  janmarqz Feb 9 '14 at 0:40
Is $XY$ the product of the summations that I gave above? –  user127292 Feb 9 '14 at 1:22
not really... what it is $XY$ is to derivate $Y$ in direction $X$. Example: $\frac{\partial}{\partial v}(v^2w\frac{\partial}{\partial w})=2vw\frac{\partial}{\partial v}+v^2w\frac{\partial}{\partial v}\frac{\partial}{\partial w}$ –  janmarqz Feb 9 '14 at 1:28
so Xg is the derivative to g in the direction X? Then what does gY mean? –  user127292 Feb 9 '14 at 1:31
$gY=\sum_j gb_j \frac \partial\partial_{x_j}$ –  janmarqz Feb 9 '14 at 1:32

Hint:

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

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

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

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

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

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