Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises - Page 150: 13


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

Work Step by Step

Given $$y=\frac{t\sin t}{ 1+t} $$ Since \begin{align*} y'&=\frac{\frac{d}{dt}\left(t\sin \left(t\right)\right)\left(1+t\right)-\frac{d}{dt}\left(1+t\right)t\sin \left(t\right)}{\left(1+t\right)^2}\\ &=\frac{\left(\sin \left(t\right)+t\cos \left(t\right)\right)\left(1+t\right)-1\cdot \:t\sin \left(t\right)}{\left(1+t\right)^2}\\ &=\frac{\sin \left(t\right)+t\cos \left(t\right)+t^2\cos \left(t\right)}{\left(1+t\right)^2} \end{align*}
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