## Calculus: Early Transcendentals 8th Edition

$J'(\theta)=2n\sec^{2}(n\theta)\tan(n\theta)$
$J(\theta)=\tan^{2}(n\theta)$ (Here, $n$ is a constant) Differentiate using the chain rule: $J'(\theta)=2\tan(n\theta)[\tan(n\theta)]'=...$ Apply the chain rule one more time to find $[\tan(n\theta)]'$ $...=2\tan(n\theta)[\sec^{2}(n\theta)](n\theta)'=2n\sec^{2}(n\theta)\tan(n\theta)$