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: 59

Answer

$f'(t)=6\pi\sec^2{(\pi t-1)}\tan{(\pi t-1)}.$

Work Step by Step

$u=\sec{(\pi t-1)}$; $\dfrac{du}{dt}=\pi\sec{(\pi t-1)}\tan{(\pi t-1)}$ $\dfrac{d}{du}f(u)=6u.$ $\dfrac{d}{dt}f(t)=\dfrac{d}{du}f(u)\times\dfrac{du}{dt}$ $=(6\sec{(\pi t-1)})(\pi\sec{(\pi t-1)}\tan{(\pi t-1)})$ $=6\pi\sec^2{(\pi t-1)}\tan{(\pi t-1)}.$
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