A Variational derivative of mean curvature

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Hello Physics Forums,

I have a simple parametric surface in R3 <x,y,z(x,y)>. I've calculated the the usual mean curvature:

H= ((1+hx^2)hyy-2hxhyhxy+(1+hy^2)hxx)/(1+hx^2+hy^2)^3/2

I needed to take the variational derivative of this expression. Since it has second order spatial derivatives the equation is:

d/dh - d/dx d/dhx - d/dy d/dhy + d^2/dx^2 d/dhxx + d^2/dxdy d/hxy +d^2/dyy d/hyy

To simplify the math I grouped the terms by d/dx and d/dy and I found that :
-dH/dhx + d/dx dH/dhxx + 1/2 d/dy dH/dhxy = 0
and
-dH/dhy + d/dy dH/dhyy + 1/2 d/dx dH/dhxy = 0

I've encountered this in my work by brute forcing it with Mathematica. However, it seems like there might be a more fundamental or elegant way of getting the result. Is there some well known reason for this? I'm an engineer working in material science. I've never taken a course on variational calculus or differntial geometry so please forgie me if this is obvious.
 
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Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
 

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