Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 20


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