## Finite Math and Applied Calculus (6th Edition)

$f^{\prime}(x)=3(x^{3}-x)^{2}\cdot(3x^{2}-1)$
Let $u(x)=x^{3},\qquad v(x)=x^{3}-x$ $\displaystyle \frac{du}{dx}=3x^{2}, \quad \frac{dv}{dx}=3x^{2}-1$ Then,$\quad y=f(x)=u(v(x))$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$ $\displaystyle \frac{du}{dv}=3v^{2}=3(x^{3}-x)^{2}$ $f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=3(x^{3}-x)^{2}\cdot(3x^{2}-1)$ $f^{\prime}(x)=3(x^{3}-x)^{2}\cdot(3x^{2}-1)$