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

Svar:

# Dy / dx = secx ^ 3 ((cos2x) / sqrt (sin2x) + 3x ^ 2tanx ^ 3sqrt (sin2x)) #

Forklaring:

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 = (secx ^ 3cos2x) / sqrt (sin2x) + 3x ^ 2secx ^ 3tanx ^ 3sqrt (sin2x) #

# Dy / dx = secx ^ 3 ((cos2x) / sqrt (sin2x) + 3x ^ 2tanx ^ 3sqrt (sin2x)) #

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

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

- (3 (x + 1)) / (2 (2x-1) ^ 2 sqrt ((x + 1) / (2x-1)) f (x) = u ^ nf '(x) = n xx du) / dx xxu ^ (n-1) I dette tilfellet: sqrt ((x + 1) / (2x-1)) = ((x + 1) / (2x-1)) ^ n = 1/2, u = (x + 1) / (2x-1) d / dx = 1/2 xx (1xx (2x-1) - 2xx (x + 1)) / xx (x x 1) / (2x-1)) ^ (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 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)))