So I'm working through Needham's Visual Differential Geometry and Forms, and what is suggested as a simple exercise is to reduce the general Gaussian curvature formula
$$K = -\frac{1}{AB}\left(\partial_{v}\frac{\partial_{v}A}{B} + \partial_{u}\frac{\partial_{u}B}{A}\right)$$
to
$$K = -\frac{\nabla^2ln\Lambda}{\Lambda^2}$$
when the metric is conformal, i.e.
$$ds^2 = \Lambda^2(du^2 + dv^2)$$
(Pages 40-41. Note, Needham uses $A^2 = E$, $B^2 = F$ in his version of the general curvature, compared to the more common formulation.) I'm sure it is simple but I am just not seeing it. Obviously, we can immediately get
$$K = -\frac{1}{\Lambda^2}\left(\partial_{v}\frac{\partial_{v}\Lambda}{\Lambda} + \partial_{u}\frac{\partial_{u}\Lambda}{\Lambda}\right)$$
with the obvious substitution, but I have no idea where to take it from there, and in particular no idea where the $ln$ would come from. I'm understanding that $\Lambda$ depends on $u, v$ essentially arbitrarily, is that my error somehow? But if it is a constant then $K$ collapses to $0$, so I am confused.
Doesn't $\dfrac{f'(x)}{f(x)}$ remind you of $(\ln f)'(x)$? This is all that's going on, but with partial derivatives.