University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 151: 26

Answer

$r' = tan^{2}\theta + cos\theta +1$ $r' = sec^{2}\theta + cos\theta$

Work Step by Step

$r = (1+sec\theta)sin\theta$ $r' = (sec\theta.tan\theta)sin\theta + (1+sec\theta)cos\theta$ $r' = tan^{2}\theta + cos\theta +1$ $r' = sec^{2}\theta + cos\theta$
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