Kako najdete derivat sinx / (1 + cosx)?

Odgovor:

# 1 / (cosx + 1) #

Pojasnilo:

#f (x) = sinx / (cosx + 1) #

#f '(x) = (sinx / (cosx + 1))' #

Izpelj iz #f (x) / g (x) # z uporabo Quotient Rule je

# (f '(x) g (x) -f (x) g' (x)) / g ^ 2 (x) #

tako je v našem primeru

#f '(x) = ((sinx)' (cosx + 1) -sinx (cosx + 1) ') / (cosx + 1) ^ 2 # #=#

# (cosx (cosx + 1) + sin ^ 2x) / (cosx + 1) ^ 2 # #=#

# (barva (modra) (cos ^ 2x) + cosx + barva (modra) (sin ^ 2x)) / (cosx + 1) ^ 2 # #=#

#cancel ((cosx + barva (modra) (1))) / (cosx + 1) ^ prekliči (2) # #=#

# 1 / (cosx + 1) #

Odgovor:

# 1 / 2sec ^ 2 (x / 2) ali 1 / (1 + cosx) #.

Pojasnilo:

Imamo, # sinx / (1 + cosx) #, # = {2sin (x / 2) cos (x / 2)} / {2cos ^ 2 (x / 2)} #,

# = tan (x / 2) #.

# "Zato," d / dx {sinx / (1 + cosx)} #, # = d / dx {tan (x / 2)} #, # = sek ^ 2 (x / 2) * d / dx {x / 2} …… "pravilo verige" #, # = sek ^ 2 (x / 2) * 1/2 #, # = 1 / 2sec ^ 2 (x / 2) ali #

# = 1 / (2cos ^ 2 (x / 2)) #, # = 1 / (1 + cosx) #.