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: 16

Answer

$f'(t)=-te^{t}\csc^{2}t+te^{t}\cot t+e^{t}\cot t$

Work Step by Step

$f(t)=te^{t}\cot t$ This is a product of three functions. To differentiate it, we will apply the product rule taking $te^{t}$ as one function. Then we will need to use the product rule a second time to finish the problem. $f'(t)=(te^{t})(\cot t)'+(\cot t)(te^{t})'=...$ Use the product rule to differentiate $te^{t}$: $...=(te^{t})(-\csc^{2}t)+(\cot t)[(t)(e^{t})'+(e^{t})(t)']=...$ $...=-te^{t}\csc^{2}t+\cot t(te^{t}+e^{t})=-te^{t}\csc^{2}t+te^{t}\cot t+e^{t}\cot t$
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