Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.3 Differentiation Rules - Exercises 3.3: 28

Answer

The Derivative is: $y'=\frac{-6x^2+12}{(x-1)^2(x-2)^2}$

Work Step by Step

$y=\frac{(x+1)(x+2)}{(x-1)(x-2)}$ Using Quotient Rule to find the Derivative: $y'=\frac{f′(x)⋅g(x)−f(x)⋅g′(x)}{g^2(x)}$ $y'=\frac{[(x^{1-1}+0)(x+2)+(x+1)(x^{1-1}+0)](x-1)(x-2)-(x+1)(x+2)[(x^{1-1}-0)(x-2)+(x-1)(x^{1-1}-0)]}{[(x-1)(x-2)]^2}$ $y'=\frac{[x+2+x+1](x-1)(x-2)-[x-2+x-1](x+1)(x+2)}{[(x-1)(x-2)]^2}$ $y'=\frac{[2x+3](x-1)(x-2)-[2x-3](x+1)(x+2)}{[(x-1)(x-2)]^2}$ $y'=\frac{2x^3-3x^2-5x+6-2x^3-3x^2+5x+6}{[(x-1)(x-2)]^2}$ $y'=\frac{-6x^2+12}{[(x-1)(x-2)]^2}$ $y'=\frac{-6x^2+12}{(x-1)^2(x-2)^2}$
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