Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 42


$$ y''=2\sin t+4t \cos t-t^2\sin t,$$ $$ y'''=6\cos t-6t\sin t-t^2\cos t.$$

Work Step by Step

Since $ y=t^2\sin t $, then we have $ y'=2t \sin t+t^2\cos t=t(2\sin t+t\cos t)$ and by the product rule, we get $$ y''= 2\sin t+t\cos t+t(3\cos t-t\sin t)=2\sin t+4t \cos t-t^2\sin t,$$ and again by using the product rule, we have $$ y'''=2\cos t+4 \cos t-4t\sin t-2t\sin t-t^2\cos t=6\cos t-6t\sin t-t^2\cos t.$$
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