Small Gradient Identity Proof

**mmmboh** · Feb9-10, 11:26 AM

Hi, I was asked to prove this identity, I found the determinants for both the left and the right side, and now I basically have to prove that (d/dy)(f(dg/dz))=(df/dy)(dg/dz), the d's are actual partials though. Can anyone give me an idea on how to prove this?

Thanks.

**LCKurtz** · Feb9-10, 11:43 AM

Originally Posted by mmmboh

Hi, I was asked to prove this identity, I found the determinants for both the left and the right side, and now I basically have to prove that (d/dy)(f(dg/dz))=(df/dy)(dg/dz), the d's are actual partials though. Can anyone give me an idea on how to prove this?

Thanks.

You can use the subscript button and subscripts to write partial derivatives. What you are asking to show is:

(fg_z)_y = f_yg_z

That's going to be difficult since the product rule gives;

(fg_z)_y = f_yg_z+ fg_zy

Hard to see where your error is if you don't show your work.

**mmmboh** · Feb9-10, 11:59 AM

$LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ x $LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ y $LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ z

f $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ x f $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ y f $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z

$LaTeX Code: \\partial$ f/ $LaTeX Code: \\partial$ x $LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ y $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z

$LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ x $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ y $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z

Those are my two determinants. I did the cross product, and since there are so many terms I will just write examples, for the first determinant my first term is ( $LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ y)(f $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z), and my first term for the second determinant is ( $LaTeX Code: \\partial$ f/ $LaTeX Code: \\partial$ y)( $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z)

So unless what I did was wrong those would have to be equal

**LCKurtz** · Feb9-10, 12:23 PM

Originally Posted by mmmboh

$LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ x $LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ y $LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ z

f $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ x f $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ y f $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z

$LaTeX Code: \\partial$ f/ $LaTeX Code: \\partial$ x $LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ y $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z

$LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ x $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ y $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z

Those are my two determinants. I did the cross product, and since there are so many terms I will just write examples, for the first determinant my first term is ( $LaTeX Code: \\partial$ / $LaTeX Code: \\partial$ y)(f $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z),

You need to expand that partial using the product rule, which will give you two terms. And where is the second term in the expansion of the determinant? You only have half of it.

and my first term for the second determinant is ( $LaTeX Code: \\partial$ f/ $LaTeX Code: \\partial$ y)( $LaTeX Code: \\partial$ g/ $LaTeX Code: \\partial$ z)

And you only have half of that determinant too.

**mmmboh** · Feb9-10, 12:32 PM

Ok the first term in the expansion of the first determinant is (fg_z)_y-(fg_y)_z, and the first term in the expansion for the second determinant is f_yg_z-g_yf_z.

When I expand the first expansion of the first determinant I get f_yg_z+fg_zy-f_zg_y-fg_yz...ah and the zy are interchangeable and so you can cancel two terms out, giving you f_yg_z-f_zg_y which equals the first expression of the second determinant! Thanks!

**LCKurtz** · Feb9-10, 12:38 PM

You're welcome.