Calculus
posted by Kate .
Differentiate with respect to theta
Y=((sin3theta)(tan3theta))^12

y = (sin3θ tan3θ)^12
y' = 12(sin3θ tan3θ)^11 (3cos3θtan3θ + 3sin3θsec^2 3θ)
= 108(sin3θtan3θ)^11(sin3θ+sin3θ(1+tan^2 3θ))
= 108(sin3θtan3θ)^11 (2sin3θ + sin3θtan^2 3θ)
or other ways
or, expanding first,
y = sin^12 3θ * tan^12 3θ
y' = 12sin^11 3θ tan^12 3θ (3cos 3θ) + 12sin^12 3θ tan^11 3θ (3sec^2 3θ)
which I'm sure massages out to the same.