## Proving div(F X G) = G·curl(F) - F·curl(G)

1. The problem statement, all variables and given/known data

If F = F1i + F2j + F3k and G = G1i + G2j + G3k are differentiable vector functions of (x,y,z) prove that div(F X G) = G·curl(F) - F·curl(G)

3. The attempt at a solution

If computed both sides of the equation, but they are not the same. My left hand side is
div(F X G) = (∂/∂x)(F2G3-G2F3) + (∂/∂y)(G1F3-F1G3) + (∂/∂z)(F1G2-F2G1)

My right hand side is
G·curl(F) - F·curl(G) = (∂/∂x)(2F2G3-2G2F3) + (∂/∂y)(2G1F3-2F1G3) + (∂/∂z)(2F1G2-2F2G1)

This weird two has appeared out of nowhere! I've checked over many, many times and still can't spot any arithmatic mistakes. I'll break it down below and hopefully somebody may be able to point out a mistake.

G·curl(F) = G1F3(∂/∂y) - G1F2(∂/∂z) - G2F3(∂/∂x) + G2F1(∂/∂z) + G3F2(∂/∂x) - G3F1(∂/∂y)

F·curl(G) = F1G3(∂/∂y) - F1G2(∂/∂z) - F2G3(∂/∂x) + F2G1(∂/∂z) + F3G2(∂/∂x) - G1F3(∂/∂y)
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 Recognitions: Gold Member Science Advisor Staff Emeritus You seem to have added rather than subtracted G·curl(F) and F·curl(G)!
 That's what it appears like, but I checked that as well. for instance, lets just group the ∂/∂x terms. From G·curl(F) we have: ∂/∂x (G3F2 - G2F3) and from F·curl(G) we have: ∂/∂x (F3G2- F2G3) Then G·curl(F) - F·curl(G) = [∂/∂x (G3F2 - G2F3)] - [∂/∂x (F3G2- F2G3)] = ∂/∂x (2F2G3 - 2F3G2) as shown above!

## Proving div(F X G) = G·curl(F) - F·curl(G)

bump!
 Recognitions: Homework Help Science Advisor From G.curl(F) I get G3*d/dx(F2)-G2*d/dx(F3). That's different from your result. Why? G.curl(F) shouldn't have ANY derivatives of G, right?
 ok, I've taken that onboard. I am still struggling though. My left hand side remains the same, div(F X G) = (∂/∂x)(F2G3-G2F3) + (∂/∂y)(G1F3-F1G3) + (∂/∂z)(F1G2-F2G1) Now taking into account the help from Dick, my RHS G·curl(F) - F·curl(G) now becomes G1(∂/∂y)F3 - G1(∂/∂z)F2 - G2(∂/∂x)F3 + G2(∂/∂z)F1 + G3(∂/∂x)F2 - G3(∂/∂y)F1 - [F1(∂/∂y)G3 - F1(∂/∂z)G2 - F2(∂/∂x)G3 + F2(∂/∂z)G1 + F3(∂/∂x)G2 - G1(∂/∂y)F3] My quesiton now, is how do I make them equal? I am thinking of product rule. On the right track?
 Recognitions: Homework Help Science Advisor Yes, product rule! Product rule!
 Done, thank you!