Svar:
#f '(x) = (1 / (ln ((x + 4) / (ln (x ^ 2 + 4)))) ((1) / ((x + 4))). ^ 2 + 4) (ln (x ^ 2 + 4)) - (2x ^ 2 + 4x)) / ((x ^ 2 + 4) (ln (x ^ 2 + 4)))) #
Forklaring:
#f '(x) = (1 / (ln ((x + 4) / (ln (x ^ 2 + 4)))) (1 / ((x + 4) / (ln (x ^ 2 + 4)))) (((1) (ln (x ^ 2 + 4)) -. (x + 4) (1) / ((x ^ 2 + 4)) (2x)) / ((ln (x ^ 2 + 4))) ^ 2) #
(ln (x + 4) / (ln (x ^ 2 + 4)))) (ln (x ^ 2 + 4) / ((x + 4))). ((ln (x ^ 2 + 4) - (2x ^ 2 + 4x) / ((x ^ 2 + 4))) / ((ln (x ^ 2 + 4))) ^ 2) #
#f '(x) = (1 / (ln (x + 4) / (ln (x ^ 2 + 4))))) (ln (x ^ 2 + 4)) / ((x + 4))) (((x ^ 2 + 4) (ln (x ^ 2 + 4)) -. (2x ^ 2 + 4x)) / ((x ^ 2 + 4) (ln (x ^ 2 + 4)) ^ annullere (2))) #
#f '(x) = (1 / (ln ((x + 4) / (ln (x ^ 2 + 4)))) ((1) / ((x + 4))). ^ 2 + 4) (ln (x ^ 2 + 4)) - (2x ^ 2 + 4x)) / ((x ^ 2 + 4) (ln (x ^ 2 + 4)))) #
