Thomas' Calculus 13th Edition

$y'=(6θ+3secθtanθ)(θ^2+secθ+1)^2$
Take the derivative of the equation, and apply Chain Rule to the inner terms: $y'=3(θ^2+secθ+1)^2\times(2θ+secθtanθ)$ $=(6θ+3secθtanθ)(θ^2+secθ+1)^2$