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: 54

Answer

$\frac{ -\sqrt[3]{2^x}}{\ln(2)}2^{x} $

Work Step by Step

Given $$y=\int_{2^{x}}^{1} \sqrt[3]{t} d t$$ Since \begin{aligned} \frac{dy}{dx}&= \frac{d}{dx} \int_{2^{x}}^{1} \sqrt[3]{t} d t\\ &=- \frac{d}{dx} \int_{1}^{2^{x}} \sqrt[3]{t} d t\\ &= -\sqrt[3]{2^x}\frac{d}{dx}(2^x)\\ &=\frac{ -\sqrt[3]{2^x}}{\ln(2)}2^{x} \end{aligned}
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