Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 32

Answer

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

Work Step by Step

$y=\frac{1}{t^3+2t^2-1}$ Use the quotient rule. $y'=\frac{(t^3+2t^2-1)(0)-(1)(3t^2+4t)}{(t^3+2t^2-1)^2}$ Distribute and simplify. $y'=\frac{-3t^2-4t}{(t^3+2t^2-1)^2}$ Factor. $y'=\frac{t(-3t-4)}{(t^3+2t^2-1)^2}$
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