Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 279: 91



Work Step by Step

By making use of (FTC II), we have $$ H'(x)=-\frac{d}{d x} \int_{9}^{4x^2} \frac{1}{t} dt =-\left(\frac{d}{d u} \int_{9}^{u} \frac{1}{t} dt\right)\frac{du}{dx}=-\frac{1}{4x^2} ( 8 x)\\ = -\frac{2}{x}. $$ Hence, $$H'(1)=-2.$$
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