Answer
$y'=\sec^{3}\theta+\sec\theta\tan^{2}\theta$
Work Step by Step
$y=\sec\theta\tan\theta$
Differentiate using the product rule:
$y'=(\sec\theta)(\tan\theta)'+(\tan\theta)(\sec\theta)'=...$
$...=(\sec\theta)(\sec^{2}\theta)+(\tan\theta)(\sec\theta\tan\theta)=...$
$...=\sec^{3}\theta+\sec\theta\tan^{2}\theta$