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

Differentiate with respect to theta

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

- Calculus -
**Steve**, Sunday, February 10, 2013 at 7:31pm
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.

## Answer this Question

## Related Questions

- maths - show that sin3theta-cos2theta =(1-sin theta)(4sin squared theta + 2sin ...
- maths - show that sin3theta-cos2theta =(1-sin theta)(4sin squared theta + 2sin ...
- calculus - a billboard 20ft tall is located on top of a building with its lower ...
- related rates with work shown - this is a relate rates problem in calculus in a ...
- trig math - Prove that tan(theta)tan(60+theta)=tan3theta
- Calculus - Determine the intervals on which the function is concave up and ...
- Calculus - I wanted to confirm that I solved these problems correctly (we had to...
- calculus - the equation V=U^5=7sqrtW-WZ relates the variables, V,U,W,Z. ...
- Calculus - Differentiate each function with respect to x: F(x)=x/(x+1)^1/2
- calculus - Differentiate with respect to x (3x^2-5ax+a^2)^4

More Related Questions