Odgovor:
#f '(x) = (1 / (ln ((x + 4) / (ln (x ^ 2 + 4))))) ((1) / ((x + 4))). (((x ^ 2 + 4) (ln (x ^ 2 + 4)) - (2 x ^ 2 + 4x)) / ((x ^ 2 + 4) (ln (x ^ 2 + 4)))) *
Obrazloženje:
#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)) (2 x)) / ((ln (x ^ 2 + 4))) ^ 2) #
#f '(x) = (1 / (ln ((x + 4) / (ln (x ^ 2 + 4))))) (ln (x ^ 2 + 4) / ((x + 4))). ((ln (x ^ 2 + 4) - (2 x ^ 2 + 4x) / ((x ^ 2 + 4))) / ((ln (x ^ 2 + 4))) ^ 2) *
#f '(x) = (1 / (ln ((x + 4) / (ln (x ^ 2 + 4))))) (poništi (ln (x ^ 2 + 4)) / ((x + 4)))) (((x ^ 2 + 4) (ln (x ^ 2 + 4).) - (2 x ^ 2 + 4x)) / ((x ^ 2 + 4) (ln (x ^ 2 + 4)) ^ otkazivanje (2))) #
#f '(x) = (1 / (ln ((x + 4) / (ln (x ^ 2 + 4))))) ((1) / ((x + 4))). (((x ^ 2 + 4) (ln (x ^ 2 + 4)) - (2 x ^ 2 + 4x)) / ((x ^ 2 + 4) (ln (x ^ 2 + 4)))) *
