Answer
$f^{\prime}(x)=8x+4$
Work Step by Step
Let $u(x)=x^{2},\qquad v(x)=2x+1$
Then,$\quad y=f(x)=u(v(x))$ and
$f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dx}=2x, \quad \frac{du}{dv}=2v=2(2x+1)$
$\displaystyle \frac{dv}{dx}=2$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=2(2x+1)\cdot 2=4(2x+1)$
$f^{\prime}(x)=8x+4$
