Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 196: 13


$y'=\dfrac{t^{2}\cos t+t\cos t+\sin t}{(1+t)^{2}}$

Work Step by Step

$y=\dfrac{t\sin t}{1+t}$ Differentiate using the quotient rule: $y'=\dfrac{(1+t)(t\sin t)'-(t\sin t)(1+t)'}{(1+t)^{2}}=...$ Use the product rule to find $(t\sin t)'$: $...=\dfrac{(1+t)[(t)(\sin t)'+(\sin t)(t)']-(t\sin t)(1)}{(1+t)^{2}}=...$ $...=\dfrac{(1+t)(t\cos t+\sin t)-t\sin t}{(1+t)^{2}}=...$ Simplify: $...=\dfrac{t\cos t+\sin t+t^{2}\cos t+t\sin t-t\sin t}{(1+t)^{2}}=...$ $...=\dfrac{t^{2}\cos t+t\cos t+\sin t}{(1+t)^{2}}$
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.