Answer
$\frac{-3x^4+8x^3+3}{(x^4+x+1)^2}$
Work Step by Step
According to the quotient rule,
$$\frac{d}{dx}\frac{x-2}{x^4+x+1}=\frac{(x^4+x+1)\frac{d}{dx}(x-2)-(x-2)\frac{d}{dx}(x^4+x+1)}{(x^4+x+1)^2}$$
$$=\frac{(x^4+x+1)(1)-(x-2)(4x^3+1)}{(x^4+x+1)^2}$$
Simplifying,
$$=\frac{(x^4+x+1)-(4x^4-8x^3+x-2)}{(x^4+x+1)^2}$$
$$=\frac{-3x^4+8x^3+3}{(x^4+x+1)^2}$$
