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: 20

Answer

$f' = \frac{1}{(t+2)^2}$

Work Step by Step

$f = \frac{t^{2} - 1}{t^{2}+t-2}$ $f' = \frac{(t^{2}+t-2)(2t) - (t^{2}-1)(2t+1)}{(t^{2}+t-2)^{2}}$ $f' = \frac{2t^{3}+2t^{2}-4t-2t^{3}-t^{2}+2t+1}{(t^{2}+t-2)^{2}}$ $f' = \frac{t^{2}-2t+1}{(t^{2}+t-2)^{2}}$ $f' = \frac{(t-1)(t-1)}{((t-1)(t+2))^2}$ $f' = \frac{1}{(t+2)^2}$
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