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: 28



Work Step by Step

Given that $p=(1+csc q)cosq$ $\implies p=cosq+cscq\space cosq$ $p=cosq+cosq.\frac{1}{sinq}$ (because cscq and sinq are reciprocals) $p=cosq+cotq$ Now differentiating both sides, we obtain: $p^{\prime}=-sinq-csc^2q$
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