I am reading some notes on Ricci flow by Peter Topping where the author introduces what he calls the 'classical entropy' for a function $u = e^{-f(t)}$:
$$N= \int_M u \log(u)dV.$$
Using the fact that $\frac{\partial}{\partial t}dV = -R \:dV$ we can then differentiate this under the integral to get:
$$ \begin{split} -\frac{dN}{dt} &= \int \left( (1 + \log u)\frac{\partial u}{\partial t} - Ru\log u \right)dV \\ &= - \int \left((1 + \log u)(\Delta u - Ru) + Ru \log u \right)dV \\ &= \int \left(\frac{|\nabla u|^2}{u} + Ru \right) dV \end{split} $$
In the case of surface entropy where we have $R$ I know how to do it as there are identities which can be used, but in this case $u$ is just a function, so I am not sure how the computation follows.