Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 141: 29



Work Step by Step

Using quotient rule, we have $\frac{dp}{dq}= \frac{[(sin q+cos q)'\times cos q]-[(cosq)'\times(sin q+cos q)]}{cos^{2}q}$ $= \frac{[(cosq-sinq)cosq]-[-sin q(sinq+cosq)]}{cos^{2}q}$ $=\frac{[cos^{2}q-sin q cos q]-[-sin^{2}q-sinqcosq]}{cos^{2}q}$ $=\frac{cos^{2}q+sin^{2}q}{cos^{2}q}$= $1+tan^{2}q=sec^{2}q$
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