Paano mo naiiba ang ibinigay na y = (secx ^ 3) sqrt (sin2x)?

Paano mo naiiba ang ibinigay na y = (secx ^ 3) sqrt (sin2x)?
Anonim

Sagot:

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

Paliwanag:

Meron kami # y = uv # kung saan # u # at # v # ay parehong mga tungkulin ng # 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)) #