Characterizing the derivative and the entropy function by the Leibniz rule

Consider an operator T:C¹(R{double-struck})→C(R{double-struck}) satisfying the Leibniz rule functional equation. T(f→g)=(Tf)→g+f→(Tg),f,g→C1(R{double-struck}). We prove that all solution operators T have the form. Tf(x)=c(x)f'(x)+d(x)f(x)ln|f(x)|,f∈C¹(R{double-struck}),x∈R{double-struck} where c,d∈C(R{double-struck}) are suitable continuous functions. If T acts on the smaller space C^k(R) for some k{greater-than above slanted equal above less-than above slanted equal}2, there are no further solutions. If T maps all of C(R{double-struck}) into C(R{double-struck}), c=0 and we only have the entropy function cfln|f| solution. We also consider the case of C¹-functions f:R{double-struck}ⁿ→R{double-struck}. More generally, if T:C¹(R{double-struck})→C¹(R{double-struck}) and A₁,A₂:C(R{double-struck})→C(R{double-struck}) are operators satisfying the generalized Leibniz rule equation. T(f·g)=(Tf)·(A1g)+(A2f)·(Tg),f,g∈C1(R), and some weak additional assumptions, the operators A₁ and A₂ are of a very restricted type and any corresponding solution T has the form Tf(x)=(c(x)f′(x)|f(x)|^p(x)sgn(f(x))+d(x)ln|f(x)||f(x)|^p(x)+1){sgnf(x)}. Here c,d,p∈C¹(R{double-struck}) are continuous functions with Im(p)⊂[0,∞) and the factor {sgnf(x)} may be present or not, yielding two different solutions. If c≠0, A₁ and A₂ must be equal and are uniquely determined by T,. A1f(x)=A2f(x)=|f(x)|^p(x)+1{sgnf(x)}. In the case that c(x)=0, we show that there are two further types of solutions of the functional equation depending only on x and f(x).