The expression for a derivative at pixel i is something like (f(i+1)-f(i))/h with h=1. If f() was a smooth function we should have divisor h=n instead of h, but this does not give similar images. Usual rules don't apply for fractals!
Method: f = (counts-trans)^0.7, compute L1 norm of gradient g of f, z=1/(1+g), zz = log(1+z), then map zz onto colormap.
kfr file:
Re: 0.36638282404933657700990078461565340243024656862608454555461054041101147888980006666406287345275758434010287299657418407215918103541997853562891000027147696827066702217911711791 Im: -0.59151923802607849414783861400556610176714425538476487531150067978143485436705539260212408281317815148344411306023903214429121890278707895272872655392192276678105241936306098075 Zoom: 1.08070014922E123 Iterations: 473799 IterDiv: 1.000000 SmoothMethod: 0 ColorMethod: 5 ColorOffset: 0 Rotate: 0.000000 Ratio: 360.000000 Colors: 255,255,255,223,0,31,191,0,63,159,0,95,127,0,127,95,0,159,63,0,191,31,0,223,0,0,255,0,31,255,0,63,255,0,95,255,0,127,255,0,159,255,0,191,255,0,223,255,0,255,255,0,255,255,0,255,255,0,255,255,0,255,255,0,255,255,0,255,255,0,255,255,0,255,255, Smooth: 1 MultiColor: 0 BlendMC: 0 MultiColors: Power: 2 FractalType: 0 Slopes: 1 SlopePower: 50 SlopeRatio: 50 SlopeAngle: 45 imag: 1 real: 1 SeedR: 0 SeedI: 0 FactorAR: 1 FactorAI: 0n=1