Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 216: 53



Work Step by Step

$F(x)=xsec^{-1}(x^3)\\ F'(x)=1\times sec^{-1}(x^3)+\frac{1}{(x^3)^2\sqrt{1-\frac{1}{(x^3)^2}}}\times 3x^2\times x\\ F'(x)=sec^{-1}(x^3)+\frac{3x^3}{x^6\sqrt{1-\frac{1}{x^6}}}\\ F'(x)=sec^{-1}(x^3)+\frac{3}{x^3\sqrt{1-\frac{1}{x^6}}}\\ F'(x)=sec^{-1}(x^3)+\frac{3}{\sqrt{x^6-1}}$
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