Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 147: 81


$$0.36 \mathrm{kg} / \mathrm{yr}$$

Work Step by Step

Given $$W(t)=\left(0.14+0.115 t-0.002 t^{2}+0.000023 t^{3}\right)^{3.4}$$ Since $$W'(t)= 3.4\left(0.14+0.115 t-0.002 t^{2}+0.000023 t^{3}\right)^{2.4}\left(0.115-0.004 t+0.000069 t^{2}\right) $$ Then \begin{align*} W'(10)&= 3.4\left(0.14+0.115(10)-0.002 (10)^{2}+0.000023 (10)^{3}\right)^{2.4}\left(0.115-0.004 (10)+0.000069 (10)^{2}\right)\\ &\approx 0.36 \mathrm{kg} / \mathrm{yr} \end{align*}
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