[timmclean]

I am not a mathematician, but I don't think this is actually begging the question.

> you can't just take an arbitrary function and use this method

Actually, you can, AFAIK. The relation f(x + E) = f(x) + f'(x)E still holds.

If we try this method with a rational function:

    f(x) = (x - 1)/(x - 2)

    f(x + E)
     = (x + E - 1) / (x + E - 2)
     = (x + E - 1)(x - E - 2) / ((x + E - 2)(x - E - 2))
     = (x^2 - x E - 2 x + x E - E^2 - 2 E - x + E + 2) / (x^2 - x E - 2 x + x E - E^2 - 2 E - 2 x + 2 E + 4)
     = (x^2 - 3 x + 2 - E) / (x^2 - 4 x + 4)
     = (x^2 - 3 x + 2) / (x^2 - 4 x + 4) - (1 / (x^2 - 4 x + 4)) E
     = (x - 1) / (x - 2) - (1 / (x - 2)^2) E

so

    f'(x) = -1 / (x - 2)^2

since

    f(x + E) = f(x) + f'(x) E

Edit to add: the key idea here being that no knowledge of differentiation is needed, just tricks for manipulating expressions involving x and E until they are in normal form.