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

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

Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}\Big(\frac{3t-4}{5t+2}\Big)^{-5}$$ Following the Chain Rule: $$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{d}{dt}\Big(\frac{3t-4}{5t+2}\Big)$$ $$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{(5t+2)\frac{d}{dt}(3t-4)-(3t-4)\frac{d}{dt}(5t+2)}{(5t+2)^2}$$ $$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{3(5t+2)-5(3t-4)}{(5t+2)^2}$$ $$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{15t+6-15t+20}{(5t+2)^2}$$ $$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{26}{(5t+2)^2}$$ $$\frac{dy}{dt}=-\frac{130(3t-4)^{-6}}{(5t+2)^{-4}}$$ $$\frac{dy}{dt}=-\frac{130(5t+2)^{4}}{(3t-4)^{6}}$$
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