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: 8

Answer

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

Work Step by Step

$f(t)=\dfrac{\cot t}{e^{t}}$ Differentiate using the quotient rule: $f'(t)=\dfrac{(e^{t})(\cot t)'-(\cot t)(e^{t})'}{(e^{t})^{2}}=...$ $...=\dfrac{(e^{t})(-\csc^{2}t)-(\cot t)(e^{t})}{(e^{t})^{2}}=\dfrac{-e^{t}\csc^{2}t-e^{t}\cot t}{e^{2t}}=...$ Take out common factor $e^{t}$ and simplify: $...=\dfrac{e^{t}(-\csc^{2}t-\cot t)}{e^{2t}}=\dfrac{-\csc^{2}t-\cot t}{e^{t}}=-\dfrac{\csc^{2}t+\cot t}{e^{t}}$
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