Mean curvature of graph over its tangent plane Let $S$ be a regular surface in $\mathbb{R}^3$ and $p\in S$ a point on the surface.
By the implicit function theorem $S$ can be locally written as a graph of a function, e.g. $V\cap S = \{ (x,y,z) \in \mathbb{R}^3 : (x,y)\in U, z=f(x,y)\}$ for some open neighbourhood $V$ of $p$, open set $U\subset \mathbb{R}^2$ and some smooth function $f: U \rightarrow \mathbb{R}.$
By choosing local coordinates we can identify $U$ as part of the tangent plane of $S$ at $p$, furthermore we can set $f^{-1}(p)=(0,0)$. In this case, the mean curvature at $p$ is given by $H=\frac{f_{xx}\;\;+f_{yy}}{2}$ (average of second derivatives at $p$) and principal curvatures are $f_{xx},f_{yy}$.
Is this correct? If it is, how can one describe this more precisely than "choosing local coordinates..."? If it is not, how could I achieve a similar result (surface as graph over its tangent plane and easy formula of $H$)?
Thank you.
 A: Your formula is correct. The mean curvature can be defined as half of the trace of the shape operator (minus the differential of the unit normal vector). If the surface is the graph of a function, the normal vector field at the point $(x,y,z)$ is given by
\begin{align*}
\frac{\nabla(z-f(x,y))}{|\nabla (z-f(x,y))|} &= \frac{\nabla z- \nabla f(x,y)}{\sqrt{1+|\nabla f|^2}} \\ &= \frac{e_3- \nabla f(x,y)}{\sqrt{1+|\nabla f|^2}}\, ,
\end{align*}
with $\nabla=(\partial_x,\partial_y,\partial_z)$. Note that, $\nabla f=(\partial_x f,\partial_y f,0)$.
The trace of the differential of this unit normal vector field is nothing but the divergence of it. Therefore
\begin{align*}
2H&=-\mathrm{div}\left(\frac{e_3- \nabla f(x,y)}{\sqrt{1+|\nabla f|^2}}\right)\\ &=\frac{\Delta f}{\sqrt{1+|\nabla f|^2}}+(\nabla f(x,y)-e_3)\cdot \nabla\left(\frac{1}{\sqrt{1+|\nabla f|^2}}\right) \, .
\end{align*}
Now, if the graph parameterization is with respect to the tangent space at $(x_0,y_0,z_0)\in S$ then $\nabla f(x_0,y_0)=0$ and, therefore, $H$ reduces to
$$2H(x_0,y_0,z_0)=\Delta f(x_0,y_0) \, .
$$
This is the formula you were writing and, in general, it is true at every point of $(x,y,f(x,y))\in S$ where $\nabla f(x,y)=0$.
