## Calculus (3rd Edition)

$f'(\theta ) =\tan \theta \sec \theta+\theta \sec^3 \theta +\theta \tan^2 \theta \sec \theta .$
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 $$f'(\theta )=(\theta)' \tan \theta \sec \theta+\theta (\tan \theta)' \sec \theta+\theta \tan \theta (\sec \theta)' \\ =\tan \theta \sec \theta+\theta \sec^2 \theta \sec \theta+\theta \tan \theta \sec \theta\tan \theta\\ =\tan \theta \sec \theta+\theta \sec^3 \theta +\theta \tan^2 \theta \sec \theta .$$