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: 29

Answer

$p' = sec^{2}q$

Work Step by Step

$\displaystyle p = \frac{sinq + cosq}{cosq}$ $\displaystyle p' = \frac{cosq(cosq -sinq) + (sinq+cosq)sinq}{cos^{2}q}$ $\displaystyle p' = \frac{cos^{2}q - sinq\ cosq + sin^{2}q + sinq\ cosq}{cos^{2}q}$ $\displaystyle p' = \frac{1}{cos^{2}q}$ $\displaystyle p' = sec^{2}q$
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