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: 31

Answer

$\displaystyle p' = \frac{-sinq(q^{2}+1)+ qcosq(q^{2}-1)}{(q^{2}-1)^{2}}$

Work Step by Step

$\displaystyle p = \frac{q*sinq}{q^{2}-1}$ $\displaystyle p' = \frac{(q^{2}-1)(sinq + q*cosq) - 2q^{2}sinq}{(q^{2}-1)^{2}}$ $\displaystyle p' = \frac{q^{2}*sinq+q^{3}*cosq -sinq -qcosq - 2q^{2}sinq}{(q^{2}-1)^{2}}$ $\displaystyle p' = \frac{-q^{2}sinq+q^{3}cosq -sinq -q*cosq}{(q^{2}-1)^{2}}$ $\displaystyle p' = \frac{-sinq(q^{2}+1)+ qcosq(q^{2}-1)}{(q^{2}-1)^{2}}$
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