Odgovor:
Pojasnilo:
Kako ločite f (x) = sqrt (cote ^ (4x) z uporabo pravila verige.?
F '(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (posteljica (e ^ (4x))) ^ (- 1/2)) / 2 barva (bela) (f') (x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (cot (e ^ (4x)) f (x) = sqrt (cot (e ^ (4x))) barva (bela) (f (x)) = sqrt (g (x)) f '(x) = 1/2 * (g (x)) ^ (- 1/2) * g' (x) barva (bela) ) (f '(x)) = (g' (x) (g (x)) ^ (- 1/2)) / 2 g (x) = posteljica (e ^ (4x)) barva (bela) (g) (x)) = posteljica (h (x)) g '(x) = - h' (x) csc ^ 2 (h (x)) h (x) = e ^ (4x) barva (bela) (h ( x)) = e ^ (j (x)) h '(x) = j' (x) e ^ (j (x)) j (x) = 4x j '(x) = 4 h' (x) = 4e ^ (4x) g '(x) = - 4e ^ (4x) cs
Kako ločite sqrt ((x + 1) / (2x-1))?
- (3 (x + 1)) / (2 (2x-1) ^ 2 sqrt ((x + 1) / (2x-1)) f (x) = u ^ n f '(x) = n xx ( du) / dx xxu ^ (n-1) V tem primeru: 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)
Kako ločite f (x) = sqrt (ln (1 / sqrt (xe ^ x)) z uporabo verižnega pravila.?
Samo z verigo vedno znova. 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, to bo težko: 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)) '= = sqr