Paano mo naiibahin ang (x ^ 2 + x + 3) / sqrt (x-3) gamit ang quotient rule?

Sagot:

#h '(x) = - 3 (x + 1) / ((x-3) ^ (3/2)) #

Paliwanag:

Ang halagang panuntunan; ibinigay #f (x)! = 0 #

kung #h (x) = f (x) / g (x) #; pagkatapos # x '(x) = g (x) * f' (x) -f (x) * g '(x) / (g (x)) ^ 2 #

ibinigay #h (x) = (x ^ 2 + x + 3) / root () (x-3) #

hayaan #f (x) = x ^ 2 + x + 3 #

#color (pula) (f '(x) = 2x + 1) #

hayaan #g (x) = root () (x-3) = (x-3) ^ (1/2) #

#color (asul) (g '(x) = 1/2 (x-3) ^ (1 / 2-1) = 1/2 (x-3) ^ (- 1/2) #

# (x) = (x-3) ^ (1/2) * kulay (pula) ((2x + 1)) - kulay (asul) (1/2 (x-3) ^ (- 1 / 2)) (x ^ 2 + x + 3) / (root () (x-3) ^ 2 #

I-factor ang pinakadakilang kadahilanan # 1/2 (x-3) ^ (- 1/2) #

# x '(x) = 1/2 (x-3) ^ (- 1/2) (x-3) (2x + 1) - (x ^ 2 + x + 3) / (x-3) #

# => h '(x) = 1/2 (x ^ 2 + x-6x-3-x ^ 2 -x-3) / (x-3) ^ (3/2) #

#h '(x) = (-6x-6) / (2 (x-3) ^ (3/2)) #

#h '(x) = - 6 (x + 1) / (2 (x-3) ^ (3/2)) #

#color (pula) (h '(x) = - 3 (x + 1) / ((x-3) ^ (3/2))) # Sagot