Answer
$$ y'= -\frac{k}{3}(kx+b)^{-4/3}.$$
Work Step by Step
Recall that $(x^n)'=nx^{n-1}$
Since $ y=(kx+b)^{-1/3}$, by using the chain rule, the derivative $ y'$ is given by
$$ y'=-\frac{1}{3}(kx+b)^{-4/3}(k)=-\frac{k}{3}(kx+b)^{-4/3}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 71
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