Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 15

Answer

$$y' = \frac{3(3x+2)}{1 + (3x+2)^3}$$

Work Step by Step

$$y = \int\limits^{3x+2}_1{\frac{t}{1+t^3}}dt$$ Using FTC 1, substitute in the upper bound for t and multiply by the derivative of the upper bound. $$y' = \frac{3x+2}{1 + (3x + 2)^3} \times (3x+2)'$$ Simplify. $$y' = \frac{3x+2}{1 + (3x + 2)^3} \times 3$$ $$y' = \frac{3(3x+2)}{1 + (3x+2)^3}$$
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