Calculus (3rd Edition)

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


$$ y'= -\frac{k}{3}(kx+b)^{-4/3}.$$

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 $$ y'=-\frac{1}{3}(kx+b)^{-4/3}(k)=-\frac{k}{3}(kx+b)^{-4/3}.$$
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