Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 196: 25

Answer

a. $y = 2x$ b. Graph
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Work Step by Step

$y = 2xsinx$ Use the product rule: $y' = \frac{d(2x)}{dx} \times + 2x \times \frac{d(sinx)}{dx}$ $y' = 2sinx + 2xcosx$ Now plug $\frac{\pi}{2}$: $y' = 2sin(\frac{\pi}{2}) + 2(\frac{\pi}{2})cos(\frac{\pi}{2})$ $y' = 2 + 0$ $y' = 2$ With this answer we can find the equation of the tangent line: $y = m(x - x_{1}) + y_{1}$ $y = 2(x -\frac{\pi}{2}) + \pi$ $y = 2x -\pi + \pi$ $y = 2x$ b. To draw this graph, draw the equation of the tangent line $(y = 2x)$ and original function $(y = 2xsinx)$
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