lnK = -ΔG°/RT = -ΔH°/RT + ΔS°/R, so d(lnK)/dT = ΔH°/RT2
The two are equivalent because ΔH° and ΔS° are related to each other by ΔH° = TΔS° at the temperature at which ΔG° = 0 and K=1, so you can substitute ΔS° = (constant)*ΔH°. For equilibrium reactions, ΔH° and ΔS° are generally of the same sign - if they are of opposite sign, the equilibrium is heavily to one side or the other at all temperatures.