Kako razlikujete y = ln ((x-1) / (x ^ 2 + 1))?

Odgovor:

# Dy / dx (- x ^ 2 + 2x + 1) / ((x ^ 2 + 1) (x-1)) *

Obrazloženje:

# Y = ln ((x-1) / (x ^ 2 + 1)) *

# Y = ln (x-1) -ln (x ^ 2 + 1) #

Koristite kvocijentno pravilo logaritama

Sada se razlikujete

# dy / dx = 1 / (x-1) -1 / (x ^ 2 + 1) * d / dx (x ^ 2 + 1) #Koristi pravilo lanca

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

# dy / dx = 1 / (x-1) - (2x) / (x ^ 2 + 1) # Uzmite lcd as ((x-1) (x ^ 2 + 1)

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

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

# Dy / dx (- x ^ 2 + 2x + 1) / ((x ^ 2 + 1) (x-1)) *