In my previous post I was asking for some clarification on the derivation of gaussian curvature.
$$ \begin{align} K=&\frac{1}{\left(E G-F^2\right)^2} \left\{\left|\begin{array}{ccc} \frac12E_{vv}+F_{uv}-\frac12G_{uu} & \frac{1}{2} E_u & F_u-\frac{1}{2} E_v \\ F_v-\frac{1}{2} G_u & E & F \\ \frac{1}{2} G_v & F & G \end{array} \right|\right. \\ & \left.-\left|\begin{array}{ccc} 0 & \frac{1}{2} E_v & \frac{1}{2} G_u \\ \frac{1}{2} E_v & E & F \\ \frac{1}{2} G_u & F & G \end{array}\right|\right\} \\ & \end{align}\tag1$$
Now, I was stuck on another problem where I need to derive another form of Gaussian curvature,
$$ \begin{align} K=-\frac{1}{4\left(E G-F^2\right)}\left|\begin{array}{ccc} E & F & G \\ E_u & F_u & G_u \\ E_v & F_v & G_v \end{array}\right|-\frac{1}{2 \sqrt{E G-F^2}}\left[\frac{\partial}{\partial u}\left(\frac{G_u-F_v}{\sqrt{E G-F^2}}\right)-\frac{\partial}{\partial v}\left(\frac{F_u-E_v}{\sqrt{E G-F^2}}\right)\right] \end{align}\tag2 $$
And I didn't see from where to start. I guess from the derivative part might be a good choice,
$$ \begin{align} &\frac{\partial}{\partial u}\left(\frac{G_u-F_v}{\sqrt{E G-F^2}}\right)-\frac{\partial}{\partial v}\left(\frac{F_u-E_v}{\sqrt{E G-F^2}}\right)\\ =&\frac{2(EG-F^2)(G_{uu}-E_{vv})-(G_u-F_v)(E_uG+EG_u-2FF_u)+(F_v-E_v)(E_vG+EG_v-2FF_v)}{2\sqrt{EG-F^2}} \end{align}\tag3 $$ Now it seems not to fit any matrix format. I was wondering how to related the first determinant of the $K$ in $(1)$ with my $(3)$. Because all the second derivative are coming from that part.