Answer
$f^{\prime}(x)=4(x^{2}+2x)^{3}\cdot(2x+2)$
Work Step by Step
Let $u(x)=x^{4},\qquad v(x)=x^{2}+2x$
$\displaystyle \frac{du}{dx}=4x^{3}, \quad \frac{dv}{dx}=2x+2$
Then,$\quad y=f(x)=u(v(x))$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=4v^{3}=4(x^{2}+2x)^{3}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=4(x^{2}+2x)^{3}\cdot(2x+2)$
