[hanneshdc]

Agreed on integrals, but the derivative is relatively simple?

If f(x) = exp(x) - ln(x) then f’(x) = exp(x) - 1/x, which is representable in eml form as well.

To the overall point though, I don’t think it helps make derivatives easier though. To refactor a function to eml’s is far more work than refactoring into something that’s trivially differentiable with the product rule and chain rule.

[gus_massa]

You mean

  f'(x) = eml(x,x) + eml(1,eml(eml(1,x)),1) + eml(eml(1, exp(eml(1, 1))),-eml(1, eml(eml(1, x))),1)

and I still have to macroexpand a few

  x-y = eml(eml(1, exp(eml(1, x))), eml(y,1)) 

but I got really bored