## Calculus 10th Edition

$f'(x)=\frac{(2x^3)(2x^2+x-2)}{(x+1)^2}$
$f(x)=x^4(\frac{(x+1)-2}{(x+1)})=x^4(\frac{x-1}{x+1})=\frac{x^5-x^4}{x+1}$ Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x^5-x^4; u'(x)=5x^4-4x^3$ $v(x)=x+1; v'(x)=1$ $f'(x)=\frac{(5x^4-4x^3)(x+1)-(1)(x^5-x^4)}{(x+1)^2}=\frac{5x^5+5x^4-4x^4-4x^3-x^5+x^4}{(x+1)^2}=$ $\frac{4x^5+2x^4-4x^3}{(x+1)^2}=\frac{(2x^3)(2x^2+x-2)}{(x+1)^2}$