I am not understanding how to derive this particular expression, which relates the inexact differential of work to the exact differential of volume, $$\delta w = -PdV $$
My attempt:
Reversible work can be defined as: $$w=-\int P dV $$ First, I integrate both sides with respect to volume, $$\frac{d}{dV}(w)=-\frac{d}{dV}(\int P dV) $$ $$ \frac{dw}{dV}=-P $$ Since the differential of work is inexact: $$ \delta w=-PdV $$ Mathematically, I am unsure about my first step. Nonetheless, this was my approach.