University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises: 12

Answer

$\displaystyle y' = \frac{-1}{(1+sinx)}$

Work Step by Step

$\displaystyle y = \frac{cosx}{1+sinx}$ $\displaystyle y' = \frac{-(1+sinx)sinx - cos^{2}x}{(1+sinx)^{2}}$ $\displaystyle y' = \frac{-sinx -sin^{2}x- cos^{2}x}{(1+sinx)^{2}}$ $\displaystyle y' = \frac{-(1+sinx)}{(1+sinx)^{2}}$ $\displaystyle y' = \frac{-1}{(1+sinx)}$
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