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September 1, 2014

September 1, 2014

Posted by **Kate** on Sunday, February 10, 2013 at 7:20pm.

Y=((sin3theta)(tan3theta))^12

- Calculus -
**Steve**, Sunday, February 10, 2013 at 7:31pmy = (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.

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