Gibbs thompson effect

For a droplet of radius r in a vapour phase the Gibbs thompson effect applies:

Δμ = μ_vapour-μ_droplet = 2σM/r, where M is the molar volume of the molecules in the droplet and σ is the surface energy.

This is derived from requiring the variation in Gibbs free energy to be zero:

dG = μ_vapourdn_vapour+μ_dropletdn_droplet + σdS = 0

Now my book writes this Gibbs thompson effect as:

Δμ = Δμ₀-2σM/r

Where I assume the subscript 0 refers to the difference in chemical potentials between infinitely large phases. Although I would think it was trivial to see this was equivalent to my equation for Gibbs Thompson effect, I can't really get my head around it. Can anyone help?

Edit: It seems that in my expression the finite radius of the droplet enhances the effective supersaturation, while in the other expression the opposite is true, i.e. the effective supersaturation drops with smaller droplets. Which is the correct? I have a hard time understanding it physically. Basically if we have a smaller droplet is the effective supersaturation then less, because the droplets surface tension makes the atoms in the droplet have high chemical potential?