Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises - Page 136: 18



Work Step by Step

$s(t)=\dfrac{1}{4-5t-t^2}=(4-5t-t^2)^{-1}$ Using the Chain Rule: $u=(4-5t-t^2)$; $\dfrac{du}{dt}=(-5-2t)$ $s(u)=u^{-1};\dfrac{d}{du}s(u)=-\dfrac{1}{u^2}$ $\dfrac{d}{dt}s(t)$ $=\dfrac{d}{du}s(u)\times\dfrac{du}{dt}$ $=(-\dfrac{1}{(4-5t-t^2)^2})(-2t-5)$ $=\dfrac{2t+5}{(4-5t-t^2)}.$
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