Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 32


$y' = 1$

Work Step by Step

Quotient Rule: $y=\frac{x^2-2ax+a^2}{x-a}$ $y' = \frac{(2x-2a)(x-a)-(x^2-2ax+a^2)(1)}{(x-a)^2} = 1$ Division then Power Rule: $y=\frac{x^2-2ax+a^2}{x-a}=\frac{(x-a)^2}{x-a} = x-a$ $y' = 1$
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