University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 7


$y'=3(\theta^2+\sec \theta+1)^2(2 \theta+\sec \theta \tan \theta)$

Work Step by Step

Given that $y=(\theta^2+\sec \theta+1)^3$ Apply derivative rules of differentiation: $f(x)=p'(x)q(x)+p(x)q'(x)$ $y'=\frac{d}{dx}[(\theta^2+\sec \theta+1)^3]$ $=3(\theta^2+\sec \theta+1)^2(2 \theta+\sec \theta \tan \theta)$ Hence, $y'=3(\theta^2+\sec \theta+1)^2(2 \theta+\sec \theta \tan \theta)$
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