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: 59


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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.