The theorem is the higher-dimensional extension of differentiation under the integral sign and reduces to that expression in some cases. Suppose f is independent of y and z, and that Ω(t) is a unit square in the yz -plane and has x limits a (t) and b (t). Then Reynolds transport theorem reduces to d d t ∫ ∫ a ( t ) b ( t ) f ( x , t ) d x = ∫ ∫ a ( t ) b ( t ) ∂ ∂ f ∂ ∂ t d x + ∂ ∂ b ( t ) ∂ ∂ t f ( b ( t ) , t ) − − ∂ ∂ a ( t ) ∂ ∂ t f ( a ( t ) , t ) , which, up to swapping x and t, is the standard expression for differentiation under the integral sign.