University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises: 50

Answer

$p' = \frac{-1}{2q^{2}}$ $p'' = \frac{1}{q^{3}}$

Work Step by Step

$p = \frac{q^{2}+3}{(q-1)^{3}+(q+1)^{3}}$ $p = \frac{q^{2}+3}{q^{3}-1-3q^{2}+3q+q^{3}+1+3q^{2}+3q}$ $p = \frac{q^{2}+3}{2q(q^{2}+3)}$ $p = \frac{1}{2q}$ $p' = \frac{-1}{2q^{2}}$ $p'' = \frac{1}{q^{3}}$
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