Tanong # 36b8c

Tanong # 36b8c
Anonim

Sa pamamagitan ng pagpaparami, #H (x) = (x-sqrt {x}) (x + sqrt {x}) = x ^ 2-x #

Sa pamamagitan ng Power Rule, #H '(x) = 2x-1 #.

Umaasa ako na ito ay kapaki-pakinabang.

Kung mapapansin mo iyan #H (x) # ang pagkakaiba ng mga perpektong parisukat at pagkatapos ang problema ay mas madali.

Kung hindi mo pagkatapos ay maaari mong gamitin ang Rule ng Produkto.

#H '(x) = uv' + vu '#

# X (x) = uv = (x-sqrt (x)) (x + x (x)) = (x-x ^ (1/2)

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

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

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

(X) = x + x / (2x ^ (1/2)) - x ^ (1/2) -1 / 2 + xx / (2x ^ (1/2)) + x ^ (1 / 2) -1 / 2 #

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

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

#H '(x) = x + x-1 #

#H '(x) = 2x-1 #