Where did I go wrong in proving the identity for div(F x G)?

[.....]

I tried to prove

div(F x G) = G.(curlF) - F.curl(G)

and ended up getting the right hand side equaling twice the left hand side, with no idea where i'd gone wrong :(

can someone show me how to prove it correctly?

and, if you have time.. see if you can pick where i went wrong?

this is how I attempted to do it:

(I'm going to stop with the bolding for vectors because it's too annoying)

F x G = (f2g3 - f3g2)i + (f3g1 - f1g3)j + (f1g2 - f2g1)k

div(F x G) = d/dx(f2g3 - f3g2) + d/dy(f3g1 - f1g3) + d/dz(f1g2 - f2g1)

curlF = (df3/dy - df2/dz)i + (df1/dz - df3/dx)j + (df2/dx - df1/dy)k

G.(curlF) = d/dx(f2g3 - f3g2) + d/dy(f3g1 - g3f1) + d/dz(f1g2 - f2g1)

i can see I'm already in trouble here... this is the left hand side already...

then I used the same method for F.(curlG) which gets me the negative of G.(curlF) , so that when I take it from G.(curlF) I get twice the left hand side...

Any help will be greatly appreciated

 

[mathman]

Your expression for G.(curlF) is just wrong. Correct expression x derivative term is g3(df2/dx)-g2(df3/dx). When you work out the rest you will see the original formula is correct.

 

[.....]

...still stuck :(

I understand what you corrected me on, but once I've worked out G.(curlF) - F.(curlG) I can't see what you could do to it to make it the left hand side of the equation.

G.(curlF) - F.(curlG) = g1(df3/dy - df2/dz) + g2(df1/dz - df3/dx) + g3(df2/dx - df1/dy) - f1(dg3/dy - dg2/dz) - f2(dg1/dz - dg3/dx) - f3(dg2/dx - dg1/dy)

right?

then where do I go from there?

 

[mathman]

Let's take the d/dx term. You have
-g2df3/dx+g3df2/dx+f2dg3/dx-f3dg2/dx, which is
d(f2g3)/dx-d(f3g2)dx, which is what you want.

 

[.....]

ah... thanks a bunch...

so..

g3df2/dx + f2dg3/dx = d(f2g3)dx

Is that just something I should've known? or are there some intermediate steps there?

 

[matt grime]

it's the product rule. traditionally learning that d(uv)/dx = vdu/dx+udv/dx comes well before you get onto div and curl.