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: 14

Answer

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

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: $...=\dfrac{\cos t+(\cos t)(\dfrac{\sin t}{\cos t})-\sin t\sec^{2}t}{(1+\tan t)^{2}}=...$ $...=\dfrac{\cos t+\sin t-\sin t\sec^{2}t}{(1+\tan t)^{2}}$
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.