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: 27



Work Step by Step

As given that $p=5+\frac{1}{cotq}$ $p=5+tanq$ (because tan $\theta$ and $cot \theta$ are reciprocals) We differentiate both sides to obtain: $p^{\prime}=0+sec^2q$ $p^{\prime}=sec^2q$
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