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: 20

Answer

$\frac{2x^3-10x^2+16x-2}{(x-2)^2}$

Work Step by Step

$\left(\frac{f}{g}\right)'$ = $\left(\frac{f'g-fg'}{g^2}\right)$ $\left(\frac{x^3-4x^2+x}{x-2}\right)'$ = $\frac{(3x^2-8x+1)(x-2)-(x^3-4x^2+x)(1)}{(x-2)^2}$ = $\frac{2x^3-10x^2+16x-2}{(x-2)^2}$
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