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: 12

Answer

$y'=\dfrac{1}{1-\sin x}$

Work Step by Step

$y=\dfrac{\cos x}{1-\sin x}$ Differentiate using the quotient rule: $y'=\dfrac{(1-\sin x)(\cos x)'-(\cos x)(1-\sin x)'}{(1-\sin x)^{2}}=...$ $...=\dfrac{(1-\sin x)(-\sin x)-(\cos x)(-\cos x)}{(1-\sin x)^{2}}=...$ $...=\dfrac{-\sin x+\sin^{2}x+\cos^{2}x}{(1-\sin x)^{2}}=...$ We know $\sin^{2}x+\cos^{2}x=1$, so we substitute and simplify: $...=\dfrac{1-\sin x}{(1-\sin x)^{2}}=\dfrac{1}{1-\sin x}$
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