Дана функция f (x) = e*ln x (1+ln^2x). Найдите F' (e)

1) f (x) = e^xcosx+e^x (-sinx) = e^x (cosx-sinx)

f (-x) = e^ (-x) (cos (-x) - sin (-x)) = (cosx+sinx) / (e^x)

2) f (x) = 1/6 * ln (-2x)

f ' (x) = 1/6 * (1 / (-2x) * (-2x) '=1/6 * (-2) / (-2x=1 / (6x)

f (-x) = 1 / (6 (-x)) = - 1 / (6x)