In this case, we can construct a morphism. Since D(n) follows the product rule, we only need to find a function f of x for each prime p which at some x has a value of p and a derivative of 1. Then we can compose those functions by multiplication for all other natural numbers. f_p(x) = x + p is one such set of functions, giving us the complete function F_n(x) = Π_(p∈P(n)) x + p, where P(n) is the set of prime factors of n. Note, the product of the empty set is defined to be 1, so F_1(x) = 1.
Finally, the homomorphism between D and the derivatives of functions is that D(n) = F'_n(0), so in some sense, D really is a derivative.