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: 67


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

Work Step by Step

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