Answer
$y'=6(5-2x)^{-4}+\frac{-1}{x^2}(\frac{2}{x}+1)^{3}$
Work Step by Step
Take derivative of y and apply chain rule:
$y'=-3(5-2x)^{-4}\times-2+4\times1/8(\frac{2}{x}+1)^{3}\times\frac{-2}{x^2}=6(5-2x)^{-4}+\frac{-1}{x^2}(\frac{2}{x}+1)^{3}$
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: 28
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