Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 15

Answer

$y'==\dfrac{t^{4}-8t^{3}+6t^{2}+9}{(t^{2}-4t+3)^{2}}$

Work Step by Step

$y=\dfrac{t^{3}+3t}{t^{2}-4t+3}$ Differentiate using the quotient rule: $y'=\dfrac{(t^{2}-4t+3)(t^{3}+3t)'-(t^{3}+3t)(t^{2}-4t+3)'}{(t^{2}-4t+3)^{2}}=...$ $...=\dfrac{(t^{2}-4t+3)(3t^{2}+3)-(t^{3}+3t)(2t-4)}{(t^{2}-4t+3)^{2}}=...$ Evaluate the products and simplify: $...=\dfrac{3t^{4}+3t^{2}-12t^{3}-12t+9t^{2}+9-2t^{4}-6t^{2}+4t^{3}+12t}{(t^{2}-4t+3)^{2}}=...$ $...=\dfrac{t^{4}-8t^{3}+6t^{2}+9}{(t^{2}-4t+3)^{2}}$
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