University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 24


$\frac{dq}{dr} = \frac{2(1-r)}{3(2r-r^{2})^{\frac{2}{3}}}$

Work Step by Step

$q = \sqrt[3] u$ where, $u = (2r-r^{2})$ Now, $\frac{dq}{du} = \frac{1}{3u^{\frac{2}{3}}} $ and, $\frac{du}{dr} =2-2r= 2(1-r)$ So, $\frac{dq}{dr} = \frac{dq}{du} \times \frac{du}{dr}$ $\frac{dq}{dr} = \frac{2(1-r)}{3u^{\frac{2}{3}}}$ $\frac{dq}{dr} = \frac{2(1-r)}{3(2r-r^{2})^{\frac{2}{3}}}$
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