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: 94

Answer

$g''(\dfrac{\pi}{6})=32\sqrt{3}.$

Work Step by Step

Using the Chain Rule: $g'(t)=2\sec^2{2t}$ Using the Chain Rule: $u=\sec{2t}$; $\dfrac{du}{dt}=2\sec{2t}\tan{2t}$ $g(u)=2u^2;\dfrac{d}{du}g'(u)=4u$ $\dfrac{d}{dt}g'(t)=\dfrac{d}{du}g'(u)\times\dfrac{du}{dt}=8\sec^2{2t}\tan{2t}.$ $g''(\dfrac{\pi}{6})=8\sec^2{\dfrac{2\pi}{6}}\tan{\dfrac{2\pi}{6}}=32\sqrt{3}.$ A computer algebra system was used to verify these results.
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