Answer
$J'(\theta)=2n\sec^{2}(n\theta)\tan(n\theta)$
Work Step by Step
$J(\theta)=\tan^{2}(n\theta)$ (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 $[\tan(n\theta)]'$
$...=2\tan(n\theta)[\sec^{2}(n\theta)](n\theta)'=2n\sec^{2}(n\theta)\tan(n\theta)$
