Answer
$y'=(3x-2)^5-\frac{1}{x^3}(4-\frac{1}{2x^2})^{-2}$
Work Step by Step
Take derivative of y and apply chain rule:
$y'=1/3(3x-2)^5\times3-1(4-\frac{1}{2x^2})^{-2}\times\frac{1}{x^3}=(3x-2)^5-\frac{1}{x^3}(4-\frac{1}{2x^2})^{-2}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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 148: 27
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