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: 37

Answer

See graph and explanations.
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Work Step by Step

See graph. Step 1. With $y=sec(x)$, we have $y'=sec(x)tan(x)$ which gives the slope of tangent lines. Step 2. At $x=-\pi/3$, $y=sec(-\pi/3)=2$ and $m_1=y'=sec(-\pi/3)tan(-\pi/3)=-2\sqrt 3$, tangent line equation: $y-2=-2\sqrt 3(x+\pi/3)$ or $y=-2\sqrt 3x+-2\sqrt 3\pi/3+2$ Step 3. At $x=\pi/4$, $y=sec(\pi/4)=\sqrt 2$ and $m_2=y'=sec(\pi/4)tan(\pi/4)=\sqrt 2$, tangent line equation: $y-\sqrt 2=\sqrt 2(x-\pi/4)$ or $y=\sqrt 2x-\sqrt 2\pi/4+\sqrt 2$
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