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: 36

Answer

See graph and explanations.
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Work Step by Step

See graph. Step 1. With $y=tan(x)$, we have $y'=sec^2x$ which give the slope of tangent lines. Step 2. At $x=-\pi/3$, $y=tan(-\pi/3)=-\sqrt 3$ and $m_1=y'=sec^2(-\pi/3)=4$, tangent line equation: $y+\sqrt 3=4(x+\pi/3)$ or $y=4x+4\pi/3-\sqrt 3$ Step 3. At $x=0$, $y=tan(0)=0$ and $m_2=y'=sec^2(0)=1$, tangent line equation: $y=x$ Step 4. At $x=\pi/3$, $y=tan(\pi/3)=\sqrt 3$ and $m_3=y'=sec^2(\pi/3)=4$, tangent line equation: $y-\sqrt 3=4(x-\pi/3)$ or $y=4x-4\pi/3+\sqrt 3$
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