Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises: 9

Answer

$ g’(-2) = \frac{8}{9} $

Work Step by Step

$ g(t) = \frac{t^{2}+1}{t^{2}-1}$ $ g’(t) = \frac{(t^{2}-1)(2t)-(t^{2}+1)(2t)}{(t^{2}-1)^2} $ $ g’(t) = \frac{-4t}{(t^{2}-1)^2} $ $ g’(-2) = \frac{-4(-2)}{((-2)^2-1)^2} $ $ g’(-2) = \frac{8}{9} $
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