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: 14

Answer

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

Work Step by Step

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