Answer
$y'=(2x+2)(2x^2+4x+1)$
Work Step by Step
Given that $y=(x+1)^2(x^2+2x)$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\frac{d}{dx}[(x+1)^2(x^2+2x)]$
$=2(x+1)(x^2+2x)+(x+1)^2(2x+2)$
$=(x+1)[2(x^2+2x)+(x+1)(2x+2)$
Hence, $y'=(2x+2)(2x^2+4x+1)$
