Consider the area function, A(r) = πr^{2}. When the radius changes from r to r + dr, the area changes from πr^{2} to π(r + Δr)^{2}. That is,

ΔA = π(r + Δr)^{2} − πr^{2}
= π[r^{2} + 2r Δr + (Δr)^{2}] − πr^{2}
= 2πr Δr + π (Δr)^{2},

so that

\frac{ΔA}{Δr} = π(2r + Δr).

Therefore, in the limit that Δr approaches 0, one has

\begin{aligned}\frac{dA}{dr} & = lim_{Δr→0} π(2r + Δr) \\ & = 2πr. \end{aligned}