It is well known to geometric analyst that the scalar curvature of a Riemannian manifold can be decomposed to two parts: one part has a divergence structure and the other part consists of lower order terms. My question is for the mean curvature of a hypersurface in a Euclidean space, dose it have a similar structure? Has any author considered a decomposition of this kind? Any reference will be useful to me. Thanks in advance.

I was unable to post an image without a big song and dance. If you go to MY STUFF and open the Michigan Math J. 1991, page 256 has the formula when the hypersurface is given as the level set of a smooth function. It is just a rotated version of the graph formula, as it must be. Meanwhile, the mean curvature is just a dimension constant times the divergence of the (oriented) unit normal, where it does not matter whther you take the divergence on the manifold or extend the unit normal field off the manifold and use the ambient divergence. All the same. I do not see how you are going to separate out first and second order derivatives. Wait, I can try to typeset: $$ n H = \frac{1}{\nabla F} \; \sum_{i=1}^{n+1} \sum_{j=1}^{n+1} \left( \delta_{ij}  \frac{F_i F_j}{\nabla F^2} \right) F_{ij} $$ That actually looks correct. Good for me. Level set of the function $F$ and $H$ is the mean curvature with one of the choices of unit normal field. 

