#### Answer

$4x^3$

#### Work Step by Step

We are trying to find $\frac{d}{dx}(x-1)(x^3+x^2+x+1)$.
Use the Product Rule, $\frac{d}{dx}u(x)v(x)=u'(x)v(x)+u(x)v'(x)$.
In this case, $u(x)=x-1$, so $u'(x)=1$. Also, $v(x)=x^3+x^2+x+1$, so $v'(x)=3x^2+2x+1$.
Plugging these in, we get:
$\frac{d}{dw}(x-1)(x^3+x^2+x+1)$
$=1*(x^3+x^2+x+1)+(x-1)(3x^2+2x+1)$
$=(x^3+x^2+x+1)+(3x^3+2x^2+x)-(3x^2+2x+1)$
$=x^3+x^2+x+1+3x^3+2x^2+x-3x^2-2x-1$
$=4x^3$