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 142: 43

Answer

$y=2x-\frac{\pi}{2}+1$ and $y=2x+\frac{\pi}{2}-1$. See graph.

Work Step by Step

Step 1. Given $y=tan(x)$, we have $y'=sec^2x$ Step 2. For a tangent line parallel to the line $y=2x$, let $y'=2$, we have $sec^2x=2$ and $sec(x)=\pm\sqrt 2$ thus $x=\pm\pi/4$ in $(-\pi/2,\pi/2)$ Step 3. At $x=\pi/4$, $y=tan(\pi/4)=1$, and the tangent line equation is $y-1=2(x-\pi/4)$ or $y=2x-\frac{\pi}{2}+1$ Step 4. At $x=-\pi/4$, $y=tan(-\pi/4)=-1$, and the tangent line equation is $y+1=2(x+\pi/4)$ or $y=2x+\frac{\pi}{2}-1$ Step 5. See graph.
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