Odgovor:
#f '(x) = (x (ln (x ^ 2 + 3)) ^ (- 1/2)) / (x ^ 2 + 3) = x / ((x ^ 2 + 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (ln (x ^ 2 + 3))) #
Pojasnilo:
Dobili smo:
# y = (ln (x ^ 2 + 3)) ^ (1/2) #
# y '= 1/2 * (ln (x ^ 2 + 3)) ^ (1 / 2-1) * d / dx ln (x ^ 2 + 3) #
#y '= (ln (x ^ 2 + 3)) ^ (- 1/2) / 2 * d / dx ln (x ^ 2 + 3) #
# d / dx ln (x ^ 2 + 3) = (d / dx x ^ 2 + 3) / (x ^ 2 + 3) #
# d / dx x ^ 2 + 3 = 2x
#y '= (ln (x ^ 2 + 3)) ^ (- 1/2) / 2 * (2x) / (x ^ 2 + 3) = (x (ln (x ^ 2 + 3)) ^ (-1/2)) / (x ^ 2 + 3) = x / ((x ^ 2 + 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2) +3) sqrt (ln (x ^ 2 + 3))) #
