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: 13

Answer

$\displaystyle y' = 4tanx\ secx - cosec^{2}x$

Work Step by Step

$\displaystyle y = \frac{4}{cosx} + \frac{1}{tanx}$ $\displaystyle y' = \frac{cosx(0) + 4sinx}{cos^{2}x} + \frac{tanx(0) - sec^{2}x}{tan^{2}x}$ $\displaystyle y' = \frac{ 4sinx}{cos^{2}x} + \frac{- sec^{2}x}{tan^{2}x}$ $\displaystyle y' = 4tanx\ secx - cosec^{2}x$
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