Answer
$y'=3(\theta^2+\sec \theta+1)^2(2 \theta+\sec \theta \tan \theta)$
Work Step by Step
Given that $y=(\theta^2+\sec \theta+1)^3$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\frac{d}{dx}[(\theta^2+\sec \theta+1)^3]$
$=3(\theta^2+\sec \theta+1)^2(2 \theta+\sec \theta \tan \theta)$
Hence, $y'=3(\theta^2+\sec \theta+1)^2(2 \theta+\sec \theta \tan \theta)$
