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: 54

Answer

$g'(t)=-5\pi\sin{2\pi t}.$

Work Step by Step

Using the Chain Rule: $u=\cos{\pi t}$; $\dfrac{du}{dt}=-\pi\sin{\pi t}.$ $\dfrac{d}{du}g(u)=10u$ $\dfrac{d}{dt}g(t)=\dfrac{d}{du}g(u)\times\dfrac{du}{dt}=(-\pi\sin{\pi t})(10\cos{\pi t})$ $=-10\pi\cos{\pi t}\sin{\pi t}=-5\pi\sin{2\pi t}$
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