Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.5 Derivatives of Trigonometric Functions - 3.5 Exercises: 28

Answer

$y'=\dfrac{x+\sin x}{1+\cos x}$

Work Step by Step

$y=\dfrac{x\sin x}{1+\cos x}$ Start the differentiation process by using the quotient rule: $y'=\dfrac{(1+\cos x)(x\sin x)'-(x\sin x)(1+\cos x)'}{(1+\cos x)^{2}}=...$ Evaluate the derivatives indicated in the numerator. Use the product rule to evaluate $(x\sin x)'$: $...=\dfrac{(1+\cos x)[x(\sin x)'+(x)'\sin x]-(x\sin x)(-\sin x)}{(1+\cos x)^{2}}=...$ Evaluate the derivatives indicated in the numerator: $...=\dfrac{(1+\cos x)(x\cos x+\sin x)+x\sin^{2}x}{(1+\cos x)^{2}}=...$ Evaluate the operations indicated in the numerator and simplify: $...=\dfrac{x\cos x+\sin x+x\cos^{2}x+\sin x\cos x+x\sin^{2}x}{(1+\cos x)^{2}}=...$ $...=\dfrac{x(\sin^{2}x+\cos^{2}x+\cos x)+\sin x(1+\cos x)}{(1+\cos x)^{2}}=...$ $...=\dfrac{x(1+\cos x)+\sin x(1+\cos x)}{(1+\cos x)^{2}}=...$ $...=\dfrac{(1+\cos x)(x+\sin x)}{(1+\cos x)^{2}}=\dfrac{x+\sin x}{1+\cos x}$
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.