Kako najdete (d ^ 2y) / (dx ^ 2) za 5 = x ^ 2-2y ^ 2?

Izbrisano, ker je bilo napačno

Odgovor:

dobim # -5 / (4y ^ 3) #

Pojasnilo:

# 2y ^ 2 = x ^ 2-5 #

# 4y dy / dx = 2x #

# dy / dx = x / (2y) #

# d / dx (dy / dx) = d / dx (x / (2y)) #

# = ((1) (2y) -x (2 (dy / dx))) / (2y) ^ 2 #

# = (2y-2x (x / (2y)) / / (4y ^ 2) #

# = (y-x (x / (2y))) / (2y ^ 2) #

# = (y-x (x / (2y))) / (2y ^ 2) * (2y) / (2y) #

# = (2y ^ 2-x ^ 2) / (4y ^ 3 #

Začeli smo z # 5 = x ^ 2-2y ^ 2 #, tako smo

# 2y ^ 2-x ^ 2 = -5, ki ustvarja drugi derivat, # (d ^ 2y) / dx ^ 2 = -5 / (4y ^ 3) #

Odgovor:

#y '' = -5/4 1 / y ^ 3 #

Pojasnilo:

# 2y (x) ^ 2-x ^ 2 + 5 = 0-> d / (dx) (2y (x) ^ 2-x ^ 2 + 5) = 4y y'-2x = 0 #

# d / (dx) (2y y'-x) = 2 ((y ') ^ 2 + y y' ') - 1 = 0 # tako

#y '' = (1 / 2- (y ') ^ 2) / y # ampak # y '= 1 / 2x / y # tako

#y '' = (2y ^ 2-x ^ 2) / (4y ^ 3) = -5/4 1 / y ^ 3 #