Hvordan skiller du mellom sqrt ((x + 1) / (2x-1))?

Svar:

# - (3 (x + 1)) / (2 (2x-1) ^ 2 sqrt ((x + 1) / (2x-1)) #

Forklaring:

#f (x) = u ^ n #

#f '(x) = n xx (du) / dx xxu ^ (n-1) #

I dette tilfellet:# sqrt ((x + 1) / (2x-1)) = ((x + 1) / (2x-1)) ^ (1/2): #

#n = 1/2, u = (x + 1) / (2x-1) #

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

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

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

Hvordan finner du z-scoreene som skiller mellom 85% av fordelingen fra området i haler av standard normalfordeling?

Hvordan skiller du mellom sqrt (cos (x ^ 2 + 2)) + sqrt (cos ^ 2x + 2)?

(dy) / (dx) = (xsen (x ^ 2 + 2) + sen (x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 ) / (dx) = 1 / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2)) * sen (x ^ 2 + 2) * 2x + 2sen (x + 2) ) / (dx) = (2xsen (x ^ 2 + 2) + 2sen (x + 2)) / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dx) = (avbryt2 (xsen (x ^ 2 + 2) + sen (x + 2))) / (cancel2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) / (dx) = (xsen (x ^ 2 + 2) + sen (x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2)))

Hvordan skiller du mellom gitt y = (secx ^ 3) sqrt (sin2x)?

Dy / dx = secx ^ 3 ((cos2x) / sqrt (sin2x) + 3x ^ 2tanx ^ 3sqrt (sin2x)) Vi har y = uv hvor u og v er begge funksjonene til x. dy / dx = uv '+ vu'u = secx ^ 3 u' = 3x ^ 2secx ^ 3tanx ^ 3 v = (sin2x) ^ (1/2) v '= (sin2x) ^ (- 1/2) / 2 * d / dx [sin2x] = (sin2x) ^ (- 1/2) / 2 * 2cos2x = (cos2x) / sqrt (sin2x) dy / dx = (sekx ^ 3cos2x) / sqrt (sin2x) + 3x ^ 2secx ^ 3tanx ^ 3sqrt (sin2x) dy / dx = secx ^ 3 ((cos2x) / sqrt (sin2x) + 3x ^ 2tanx ^ 3sqrt (sin2x))