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 - Page 169: 24


$y'=-\dfrac{2\cos x}{(1+\sin x)^{2}}$

Work Step by Step

$y=\dfrac{1-\sin x}{1+\sin x}$ Start the differentiation process by using the quotient rule: $y'=\dfrac{(1+\sin x)(1-\sin x)'-(1-\sin x)(1+\sin x)'}{(1+\sin x)^{2}}=...$ Evaluate the derivatives indicated in the numerator: $...=\dfrac{-(1+\sin x)(\cos x)-(1-\sin x)(\cos x)}{(1+\sin x)^{2}}=...$ Simplify: $...=\dfrac{-\cos x-\sin x\cos x-\cos x+\sin x\cos x}{(1+\sin x)^{2}}=...$ $...=\dfrac{-2\cos x}{(1+\sin x)^{2}}=-\dfrac{2\cos x}{(1+\sin x)^{2}}$
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