#### 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:
$y'=\frac{f′(x)⋅g(x)−f(x)⋅g′(x)}{g^2(x)}$
$y'=\frac{[(x^{1-1}+0)(x+2)+(x+1)(x^{1-1}+0)](x-1)(x-2)-(x+1)(x+2)[(x^{1-1}-0)(x-2)+(x-1)(x^{1-1}-0)]}{[(x-1)(x-2)]^2}$
$y'=\frac{[x+2+x+1](x-1)(x-2)-[x-2+x-1](x+1)(x+2)}{[(x-1)(x-2)]^2}$
$y'=\frac{[2x+3](x-1)(x-2)-[2x-3](x+1)(x+2)}{[(x-1)(x-2)]^2}$
$y'=\frac{2x^3-3x^2-5x+6-2x^3-3x^2+5x+6}{[(x-1)(x-2)]^2}$
$y'=\frac{-6x^2+12}{[(x-1)(x-2)]^2}$
$y'=\frac{-6x^2+12}{(x-1)^2(x-2)^2}$