Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 10


$f'(x)=-\dfrac{2x}{3\sqrt[3] {(x^{2}-1)^{4}}}$

Work Step by Step

$f(x)=\dfrac{1}{\sqrt[3] {x^{2}-1}}$ Let's do some algebra to change the appearance of the function: $f(x)=\dfrac{1}{\sqrt[3] {x^{2}-1}}=\dfrac{1}{(x^{2}-1)^{1/3}}=(x^{2}-1)^{-1/3}$ Now, differentiate using the chain rule: $f'(x)=-\dfrac{1}{3}(x^{2}-1)^{-4/3}(x^{2}-1)'=...$ $...=-\dfrac{1}{3}(x^{2}-1)^{-4/3}(2x)=-\dfrac{2}{3}x(x^{2}-1)^{-4/3}=...$ $...=-\dfrac{2x}{3(x^{2}-1)^{4/3}}=-\dfrac{2x}{3\sqrt[3] {(x^{2}-1)^{4}}}$
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