Answer
$ f'(\theta )
=\tan \theta \sec \theta+\theta \sec^3 \theta +\theta \tan^2 \theta \sec \theta .$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\sec x)'=\sec x\tan x$.
Recall that $(\tan x)'=\sec^2 x$.
Since $ f(\theta )= \theta \tan \theta \sec \theta $, then using the product rule, the derivative is given by
$$ f'(\theta )=(\theta)' \tan \theta \sec \theta+\theta (\tan \theta)' \sec \theta+\theta \tan \theta (\sec \theta)' \\
=\tan \theta \sec \theta+\theta \sec^2 \theta \sec \theta+\theta \tan \theta \sec \theta\tan \theta\\
=\tan \theta \sec \theta+\theta \sec^3 \theta +\theta \tan^2 \theta \sec \theta .$$
