Thomas' Calculus 13th Edition

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 149: 60

Answer

$\frac{-130(5t+2)^4}{(3t-4)^6}$

Work Step by Step

y=$(\frac{3t-4}{5t+2})^{-5}$ Apply the chain rule: $\frac{dy}{dt}=-5(\frac{3t-4}{5t+2})^{-6}.\frac{(5t+2).3-(3t-4).5}{(5t+2)^2}$ =-5$(\frac{5t+2}{3t-4})^6.\frac{15t+6-15t+20}{(5t+2)^2}$ =$-5\frac{(5t+2)^6}{(3t-4)^6}.\frac{26}{(5t+2)^2}$ =$\frac{-130(5t+2)^4}{(3t-4)^6}$
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