Bagaimana Anda membedakan y = ln ((x-1) / (x ^ 2 + 1))?

Bagaimana Anda membedakan y = ln ((x-1) / (x ^ 2 + 1))?
Anonim

Menjawab:

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

Penjelasan:

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

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

Gunakan aturan hasil bagi dari logaritma

Sekarang bedakan

# dy / dx = 1 / (x-1) -1 / (x ^ 2 + 1) * d / dx (x ^ 2 + 1) #Gunakan aturan rantai

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

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

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