Calculus (3rd Edition)

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

Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 38


$$ y'= -\frac{3}{x^2}\left(1+\frac{1}{x}\right)^2 .$$

Work Step by Step

Since $ y=(1+\frac{1}{x})^3$, then by the chain rule $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, the derivative $ y'$ is given by $$ y'=3\left(1+\frac{1}{x}\right)^2\left(1+\frac{1}{x}\right)'\\=3\left(1+\frac{1}{x}\right)^2\left(-\frac{1}{x^2}\right) = -\frac{3}{x^2}\left(1+\frac{1}{x}\right)^2 .$$
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