# How does $F_j \frac{\partial F_j}{\partial x_i} = \frac{1}{2} \frac{\partial(F_j F_j)}{dx_i}$

How does $$F_j \frac{\partial F_j}{\partial x_i} = \frac{1}{2} \frac{\partial(F_j F_j)}{\partial x_i}$$

For $\mathbf{F}(x_1,x_2,x_3)$ being a continuously differentiable vector field?

• wouldn't your $dx_i$ be $\partial x _i$? – janmarqz Dec 22 '14 at 17:00
• Yes, that was just a mistake. Sorry – Mr Croutini Dec 22 '14 at 17:01

$$\frac{\partial(F_j F_j)}{\partial x_i}=F_j\frac{\partial F_j}{\partial x_i}+\frac{\partial F_j}{\partial x_i}F_j=2F_j\frac{\partial F_j}{\partial x_i}.$$
$$\partial_{x_i}F_jF_j = (\partial_{x_i}F_j)F_j + F_j\partial_{x_i}F_j = 2 F_j\partial_{x_i}F_j$$ this is true for $F_j$ having the same index. i.e. $F_j=x$ then $$x\frac{\partial x}{\partial x_i} = \frac{1}{2}\frac{\partial x^2}{\partial x_i}$$ this is different when we are trying to compute the relationship for $$F_j\frac{\partial F_i}{\partial x_i}$$ for example