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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?

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    $\begingroup$ wouldn't your $dx_i$ be $\partial x _i$? $\endgroup$ – janmarqz Dec 22 '14 at 17:00
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    $\begingroup$ Yes, that was just a mistake. Sorry $\endgroup$ – Mr Croutini Dec 22 '14 at 17:01
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By the use of Libeniz' rule:

$$\frac{\partial(F_j F_j)}{\partial x_i}=F_j\frac{\partial F_j}{\partial x_i}+\frac{\partial F_j}{\partial x_i}F_j=2F_j\frac{\partial F_j}{\partial x_i}.$$

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    $\begingroup$ Ah, that makes sense now! Thank you :) $\endgroup$ – Mr Croutini Dec 22 '14 at 17:01
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$$ \partial_{x_i}F_jF_j = (\partial_{x_i}F_j)F_j + F_j\partial_{x_i}F_j = 2 F_j\partial_{x_i}F_j $$ 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

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