Kako razlikovati sqrt ((x + 1) / (2x-1))?

Odgovor:

# - (3 (x + 1)) / (2 (2x-1) ^ 2 sqrt ((x + 1) / (2x-1)) #

Obrazloženje:

#f (x) = u ^ n #

#f '(x) = n xx (du) / dx xxu ^ (n-1) #

U ovom slučaju:# sqrt ((x + 1) / (2x-1)) = ((x + 1) / (2x-1)) ^ (1/2): #

#n = 1/2, u = (x + 1) / (2x-1) #

# d / dx = 1/2 xx (1xx (2x-1) - 2xx (x + 1)) / (2x-1) ^ 2 xx ((x + 1) / (2x-1)) ^ (1 / 2-1) #

# = 1 / 2xx (-3) / ((2x-1) ^ 2 xx ((x + 1) / (2x-1)) ^ (1 / 2-1) #

# = - (3 (x + 1)) / (2 (2x-1) ^ 2 ((x + 1) / (2x-1)) ^ (1/2) #

Što je (sqrt (5+) sqrt (3)) / (sqrt (3+) sqrt (3+) sqrt (5)) - (sqrt (5-) sqrt (3)) / (sqrt (3+) sqrt) (3) sqrt (5))?

2/7 Primamo, A = (sqrt5 + sqrt3) / (sqrt3 + sqrt3 + sqrt5) - (sqrt5-sqrt3) / (sqrt3 + sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - (sqrt5) -sqrt3) / (2sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3-sqrt5) = ((sqrt5 + sqrt3) (2sqrt3-sqrt5) - (sqrt5-sqrt3) (2sqrt3 + sqrt5) (2sqrt3 + sqrt5) / ((2sqrt15-5 + 2 * 3-sqrt15) - (2sqrt15 + 5-2 * 3-sqrt15)) / ((2sqrt3) ^ 2- (sqrt5) ^ 2) = (poništi (2sqrt15) -5 + 2 * 3kkazati (-sqrt15) - otkazati (2sqrt15) -5 + 2 * 3 + otkazati (sqrt15)) / (12-5) = ( Imajte na umu da, ako su u nazivnicima (sqrt3 + sqrt (3 + sqrt5)) i (sqrt3 + sqrt (3-sqrt5)), odgovor će biti promijenjen.

Kako razlikovati sqrt (cos (x ^ 2 + 2)) + sqrt (cos ^ 2x + 2)?

(dy) / (dx) = (xsen (x ^ 2 + 2) + sen (x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dy ) / (dx) = 1 / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) * sen (x ^ 2 + 2) * 2x + 2sn (x + 2) (dy ) / (dx) = (2xsen (x ^ 2 + 2) + 2sn (x + 2)) / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dy) / (dx) = (poništi2 (xsen (x ^ 2 + 2) + sen (x + 2))) / (cancel2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dy) / (dx) = (xsen (x ^ 2 + 2) + sen (x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2)))

Kako razlikovati f (x) = sqrt (ln (1 / sqrt (xe ^ x)) pomoću lančanog pravila.

Samo pravilo lanca iznova i iznova. f '(x) = e ^ x (1 + x) / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) f (x) = sqrt (ln (1 / sqrt (xe ^ x))) Ok, ovo će biti teško: f '(x) = (sqrt (ln (1 / sqrt (xe ^ x))))' = = 1 / (2sqrt (ln (1 / sqrt (xe ^ x)))) * (ln (1 / sqrt (xe ^ x))) '= = 1 / (2sqrt (ln (1 / sqrt (xe ^ x)))) * 1 / (1 / sqrt (xe ^ x)) (1 / sqrt (xe ^ x)) '= = 1 / (2sqrt (ln (1 / sqrt (xe ^ x)))) * sqrt (xe ^ x) (1 / sqrt (xe ^ x)) '= = sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x)))) (1 / sqrt (xe ^ x))' = = sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x)))) ((xe ^ x) ^ - (1/2)