University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 344: 124


$-\dfrac{\sin x}{1+\cos ^2 x}$

Work Step by Step

Here, $y=-f'(\sec x)$ and $y'=\dfrac{d(\sec x)}{dx} =-\dfrac{(\sec x\tan x)}{\sec^2 x+1}$ or, $=-\dfrac{(1/\cos x)(\sin x/\cos x)}{1/\cos^2 x+1}$ or, $=-\dfrac{\sin x}{1+\cos ^2 x}$
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