Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises - Page 158: 12

Answer

$-\frac{2x}{3(x^2-1)^{\frac{4}{3}}}$

Work Step by Step

Original Expression: $\frac{1}{\sqrt[3](x^2-1)}$ $u = x^2-1$ $f(u) = \frac{1}{\sqrt[3] u} = u^{-\frac{1}{3}} $ Apply the chain rule: $f’(u)u’ $ $f’(u) =-\frac{1}{3}(u)^{-\frac{4}{3}} 2x$ $f’(x) =-\frac{1}{3}(x^2-1)^{-\frac{4}{3}} 2x$ Simplify: $-\frac{2x}{3(x^2-1)^{\frac{4}{3}}}$
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