University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 50


$3\cdot(\int_{0}^{x} ({t^3+1})^{10}dt)^2 (x^3+1)^{10} $

Work Step by Step

Let's differentiate both sides and then use the chain rule to differentiate! $\frac{dy}{dx}=\frac{d}{dx}(\int_{0}^{x} ({t^3+1})^{10}dt)^3=3\cdot(\int_{0}^{x} ({t^3+1})^{10}dt)^2 (x^3+1)^{10} $
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