Exercise:
Show that if $z=f(x,y)$, then for its mean curvature $H=\frac{1}{2} \frac{g_{11}b_{22}+g_{22}b_{11}-g_{12}b_{12}}{g_{11}g_{22}-g_{12}}$
$$2H = \left( \nabla \cdot \left( \frac{\nabla f}{1 + \nabla^2 f} \right) \right) ^{\frac{1}{2}}.$$
I can not arrive at the equality.
My solution:
Left side.
The first fundamental form is $$\begin{pmatrix} 1 + f^2_{x}& f_x f_y \\ f_x f_y& 1+f_y^2\end{pmatrix},$$
the second fundamental form is $$\frac{1}{\sqrt{1+f_x^2+f_y^2}} \begin{pmatrix} f_{xx}& f_{xy} \\ f_{xy}& f_{yy} \end{pmatrix}.$$
$$ 2H =\frac{1}{\sqrt{1+f_x^2+f_y^2}} \frac{(1 + f^2_{x})(f_{yy}) + (1+f_y^2)(f_{xx}) - (f_x f_y)(f_{xy})}{(1 + f^2_{x})(1 + f^2_{y})-(f_x f_y)^2} = \\ = \frac{1}{\sqrt{1+f_x^2+f_y^2}} \frac{f_{xx}+f_{yy}+f^2_{x}f_{yy}+f^2_y f_{xx} - f_{xy}f_xf_y}{1+f_x^2+f_y^2}. $$
Right side.
$$\frac{\nabla f}{1+\nabla^2 f} = \frac{1}{1+f_x^2+f_y^2 }\begin{pmatrix} f_x \\ f_y \end{pmatrix}.$$
$$\nabla \cdot \begin{pmatrix}\frac{f_x}{1+f_x^2+f_y^2 } \\\frac{f_y}{1+f_x^2+f_y^2 }\end{pmatrix} = \frac{f_{xx}(1+f_x^2+f_y^2) - f_x(2f_x f_{xx}+2f_yf_{yx}) + f_{yy}(1+f_x^2+f_y^2)-f_y(2f_xf_{xy}+2f_yf_{yy})}{(1+f_x^2+f_y^2)^2}, $$
which, after we put this expression under square root, does not appear to be equal to the left side.
What am I doing wrong?
I may have also misread the writing on the board, and would be very grateful if someone told the correct version.
The expression on the board is a little ambiguous and actually contains an error. The right hand side should read $$ \nabla\cdot\left(\frac{\nabla f}{\sqrt{1+|\nabla f|^2}}\right) $$
The expression on the left hand side contains two errors as well. One of them you fixed in your evaluation, namely in the denominator the second term should be $g_{12}^2$. The other error is in the last term of the numerator. It's missing a factor of $2$. Combining these two corrections, the mean curvature should be $$ H = \frac{1}{2} \frac{g_{11}b_{22}+g_{22}b_{11} - \color{red}{2} g_{12}b_{12}}{g_{11}g_{22}-g_{12}^\color{red}{2}}\ . $$
With these corrections, you should be able to verify the formula.