Consider the inexact differential form, δ δ u = 2 y d x + x d y . This must be inexact by considering going to the point (1,1). If we first increase y and then increase x, then that corresponds to first integrating over y and then over x. Integrating over y first contributes ∫ ∫ 0 1 x d y | x = 0 = 0 and then integrating over x contributes ∫ ∫ 0 1 2 y d x | y = 1 = 2 . Thus, along the first path we get a value of 2. However, along the second path we get a value of ∫ ∫ 0 1 2 y d x | y = 0 + ∫ ∫ 0 1 x d y | x = 1 = 1 . We can make δ δ u an exact differential by multiplying it by x, yielding x δ δ u = 2 x y d x + x 2 d y = d ( x 2 y ) . And so x δ δ u is an exact differential.