Kako ločite f (x) = (3x ^ 3-2x ^ 2 + 5) ^ 331?

Kako ločite f (x) = (3x ^ 3-2x ^ 2 + 5) ^ 331?
Anonim

Odgovor:

# (dy) / (dx) = 331 (9x ^ 2-4x) (3x ^ 3-2x ^ 2 + 5) ^ 330 #

Pojasnilo:

Z uporabo verižnega pravila: # (dy) / (dx) = (dy) / (du) * (du) / (dx) #

V tem primeru, # y = (3x ^ 3-2x ^ 2 + 5) ^ 331 #

Let # u = 3x ^ 3-2x ^ 2 + 5 #, potem # (dy) / (du) = 331u ^ 330 # in # (du) / (dx) = 9x ^ 2-4x #

Torej # (dy) / (dx) = 331u ^ 330 * (9x ^ 2-4x) #

# = 331 (9x ^ 2-4x) (3x ^ 3-2x ^ 2 + 5) ^ 330 #