Thomas' Calculus 13th Edition

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 149: 71


$y'=\frac{-3}{x^2}(1+\frac{1}{x})^2$ $y''=\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$

Work Step by Step

y=$(1+\frac{1}{x})^3$ Apply the chain rule: $ y'=3(1+\frac{1}{x})^2(\frac{-1}{x^2})$ =$\frac{-3}{x^2}(1+\frac{1}{x})^2$ $ y''=(-\frac{3}{x^2}).\frac{d}{dx}(1+\frac{1}{x})^2.\frac{d}{dx}(-(1+\frac{1}{x}))^2.\frac{3}{x^2}$ =$\frac{-3}{x^2}(2(1+\frac{1}{x}))(-\frac{1}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$ =$\frac{6}{x^4}(1+\frac{1}{x})+\frac{6}{x^3}(1+\frac{1}{x})^2$ =$\frac{6}{x^2}(1+\frac{1}{x})(\frac{1}{x}+1+\frac{1}{x})$ =$\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$
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