Thomas' Calculus 13th Edition

The Derivative is: $y'=\frac{-4x^3-3x^2+1}{(x^2-1)^2(x^2+x+1)^2}$
$y=\frac{1}{(x^2-1)(x^2+x+1)}$ Using Quotient Rule to find the Derivative: $y'=\frac{f′(x)⋅g(x)−f(x)⋅g′(x)}{g^2(x)}$ $y'=\frac{(0)(x^2-1)(x^2+x+1)-(1)[((2)x^{2-1}-0)(x^2+x+1)+(x^2-1)((2)x^{2-1}+x^{1-1}+0)]}{[(x^2-1)(x^2+x+1)]^2}$ $y'=\frac{-[2x(x^2+x+1)+(2x+1)(x^2-1)]}{[(x^2-1)(x^2+x+1)]^2}$ $y'=\frac{-2x^3-2x^2-2x-2x^3+2x-x^2+1}{[(x^2-1)(x^2+x+1)]^2}$ $y'=\frac{-4x^3-3x^2+1}{[(x^2-1)(x^2+x+1)]^2}$ $y'=\frac{-4x^3-3x^2+1}{(x^2-1)^2(x^2+x+1)^2}$