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: 11

Answer

$\displaystyle y' = \frac{-cosec^{2}x}{(1+cotx)^{2}}$

Work Step by Step

$\displaystyle y = \frac{cotx}{1+cotx}$ $\displaystyle y' = \frac{-(1+cotx)cosec^{2}x - (-cosec^{2}x)cotx}{(1+cotx)^{2}}$ $\displaystyle y' = \frac{-cosec^{2}x -cotx\ cosec^{2}x + cotx\ cosec^{2}x}{(1+cotx)^{2}}$ $\displaystyle y' = \frac{-cosec^{2}x}{(1+cotx)^{2}}$
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