Just to throw my tuppence in. I do this differently in Infinity Fractal. I suppose it might be considered less elegant, but it works for all formulas and colouring methods.

Starting with a colouring method such as Smooth (log), Smooth (exp) or Triangle Inequality Average - for example:
Each of these methods results in a double representing a location in the palette (where the number is greater than the length of the palette it is wrapped around)
So for slope, I consider this number to represent the pixel's/orbit's height.
Now, when colouring the pixel, I check the surrounding (eight) pixels' heights to calculate how much brighter or darker this pixel should be.

Eg:

c = pix.ColourIndex;
l = c; // lowest index
h = c; // highest index
a = 0; // colour addition
m = 64; // multiplier

for each surrounding pixel (sp)
{
    sc = sp.ColourIndex;
    if (sc < l) l = sc;
    if (sc > h) h = sc;

    yy = -1 / 0 / 1 (sp position relative to p)
    xx = -1 / 0 / 1 (sp position relative to p)

    hd = c - pc;
    if (hd > 0)
    {
        if (yy > 0)
            a -= hd;
        else
            a += hd;
        if (xx > 0)
            a -= hd;
        else
            a += hd;
    }

    each element of the colour (R, G and B) are then added to by:
        a * m / (h - l)
}

Wow Gerrit,

That's one nice image. Where does one find it in the set?

Referring to https://fractalforums.org/fractal-mathematics-and-new-theories/28/mandelbrot-foam-math/1665/msg8427#msg8427, \( q=0.35i \), location -0.0523992616496 + i*0.185100952071,
magnification 308.

@mrrudewords is it working on infinity.1.0.20?

It is indeed. Use the Embossed (slope) effect with (eg) Smooth (log) colouring.

Wow   mrrudewords,

It's beautiful. Well done. I do have a question though.

Now, when colouring the pixel, I check the surrounding (eight) pixels' heights to calculate how much brighter or darker this pixel should be.

What is the advantage at looking at 8 pixels as opposed to 4 as used in the code recommended by superheal? Does this add extra processing time?

Thanks.

@mrrudewords a very striking image, well done! (that would be a contest winner  )

Thanks guys

Glad to be of service