Relationship of the Kaniadakis Distribution to d ln(p)/dkx = ln(p) r(kx)

In (1), the Kaniadakis distribution p(kx) (where x=-ei/T) is studied as a solution of the differential equation sqrt(1+kkxx) dp/dx + p = 0. Here we argue that p(kx) is the solution such that dp/dkx returns p(x) b(x). We further suggest that p(kx) = { g(kx) + kx } power 1/k so that for x>>1, one has p(kx)-> kx power 1/k. For x=0, g(kx) cannot be 0 as this is unphysical. The question becomes: How does one find g(kx)? We consider dp/dx = p yielding the Maxwell-Boltzmann solution p(x) = exp(x) for x= -ei/T. If one wishes the above distribution {} power 1/k to become the Maxwell-Boltzmann for x->0, then it is actually the function within {] which should become exp(kx). This means that one should have a differential equation apply to the argument {} of {} power 1/k, i.e. d ln(p)/dkx = ln(p(kx)) r(kx) such that r(kx) ->1 + 0*x in an x->0 expansion to create the MB distribution. This means that dg/dkx + 1 = (g+kx) * r(kx). There are two ways in which this may occur. First, dg/dxk= g/xk which means that g=xk which implies p(0)=0 which is unphysical or dg/dxk = xk/g which yields g=sqrt(1+kkxx) and r=1/g. This then satisfies the conditions above and leads to the Kanidakis distribution: p(kx) = {sqrt(1+kkxx) + kx} power 1/k.