Answer
$\displaystyle \frac{dy}{dx}=(x+1)(3x-1)$
Work Step by Step
$f(x)=x+1,\ \ \quad g(x)=x^{2}-1,\quad y=f(x)\cdot g(x)$
$f^{\prime}(x)=1\qquad g^{\prime}(x)=2x$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)\cdot g(x)]$= ... product rule ...
=$f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$
$=1\cdot(x^{2}-1)+(x+1)\cdot 2x$
$=(x+1)(x-1)+(x+1)\cdot 2x$
$=(x+1)(x-1+2x)$
$\displaystyle \frac{dy}{dx}=(x+1)(3x-1)$
