Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Practice Exercises - Page 178: 58



Work Step by Step

Step 1. Given $s=t^2+5t$ and $t=(u^2+2u)^{1/3}$, we have $\frac{ds}{dt}=2t+5$ and $\frac{dt}{du}=\frac{1}{3}(u^2+2u)^{-2/3}(2u+2)$ Step 2. $\frac{ds}{du}=\frac{ds}{dt}\frac{dt}{du}=(2t+5)(\frac{1}{3}(u^2+2u)^{-2/3}(2u+2))=(2(u^2+2u)^{1/3}+5)(\frac{1}{3}(u^2+2u)^{-2/3}(2u+2))$ Step 3. At $u=2$, $\frac{ds}{du}=(2(2^2+2(2))^{1/3}+5)(\frac{1}{3}(2^2+2(2))^{-2/3}(2(2)+2))=(2(8)^{1/3}+5)(\frac{1}{3}(8)^{-2/3}(6))=(9)(\frac{1}{12}(6))=\frac{9}{2}$
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