Responda:
#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)))) #
Explicação:
#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) #
#f '(x) = (1 / (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))))) (cancelar (ln (x ^ 2 + 4)) / ((x + 4)) (. ((x ^ 2 + 4) (ln (x ^ 2 + 4)) - (2x ^ 2 + 4x)) / ((x ^ 2 + 4) (ln (x ^ 2 + 4)) ^ cancelar (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)))) #
