Answer
$q'=\frac{2-2r}{3\sqrt[3] (2r-r^2)^2}$
Work Step by Step
Take the derivative and apply the chain rule:
$q'=1/3(2r-r^2)^{-2/3}\times(2-2r)=\frac{2-2r}{3\sqrt[3] (2r-r^2)^2}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 148: 20
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