Answer
$g'(x)=\frac{x^2+2x+2}{(x+1)^2}$
Work Step by Step
$g(x)=x^2(\frac{2(x+1)-x}{x(x+1)})=\frac{x^3+2x^2}{x^2+x}$
Using the quotient rule: $g’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=x^3+2x^2; u'(x)=3x^2+4x$
$v(x)=x^2+x; v'(x)=2x+1$
$g'(x)\frac{(3x^2+4x)(x^2+x)-(2x+1)(x^3+2x^2)}{(x(x+1))^2}=$
$\frac{(3x^4+7x^3+4x^2)-(2x^4+5x^3+2x^2)}{x^2(x+1)^2=}$
$\frac{x^2(x^2+2x+2)}{x^2(x+1)^2}=\frac{x^2+2x+2}{(x+1)^2}$
