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: 32

Answer

$\displaystyle p' = \frac{q*sec^{2}q -tanq(3q^{2}+1+qtanq)}{q^{2}secq}$

Work Step by Step

$\displaystyle p = \frac{3q+tanq}{qsecq}$ $\displaystyle p' = \frac{qsecq(3+sec^{2}q) - (3q+tanq)(secq+qsecqtanq)}{q^{2}sec^{2}q}$ $\displaystyle p' = \frac{q*sec^{3}q -3q^{2}secq*tanq - tanq*secq- qsecq*tan^{2}q}{q^{2}sec^{2}q}$ $\displaystyle p' = \frac{q*sec^{2}q -tanq(3q^{2}+1+qtanq)}{q^{2}secq}$
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