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: 14


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

Work Step by Step

$y=\dfrac{\sin t}{1+\tan t}$ Differentiate using the quotient rule: $y'=\dfrac{(1+\tan t)(\sin t)'-(\sin t)(1+\tan t)'}{(1+\tan t)^{2}}=...$ $...=\dfrac{(1+\tan t)(\cos t)-(\sin t)(\sec^{2}t)}{(1+\tan t)^{2}}=...$ Evaluate the products in the numerator: $...=\dfrac{\cos t+\cos t\tan t-\sin t\sec^{2}t}{(1+\tan t)^{2}}$ We know $\tan t=\dfrac{\sin t}{\cos t}$, so we substitute and simplify: $...=\dfrac{\cos t+(\cos t)(\dfrac{\sin t}{\cos t})-\sin t\sec^{2}t}{(1+\tan t)^{2}}=...$ $...=\dfrac{\cos t+\sin t-\sin t\sec^{2}t}{(1+\tan t)^{2}}$
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