## Calculus with Applications (10th Edition)

$\displaystyle \frac{dy}{dx}=\frac{60x^{2}}{(2x^{3}+1)^{3}}$
$y=\displaystyle \frac{-5}{(2x^{3}+1)^{2}}=-5(2x^{3}+1)^{-2}=\mathrm{k}\cdot v(x)$ Use the Constant Times a Function rule $\color{blue}{[\mathrm{k}\cdot v(x)]^{\prime}=k\cdot v^{\prime}(x)}$ For $v^{\prime}(z)$, we need the chain rule. $\left[\begin{array}{llll} & & v(x)=(2x^{3}+1)^{-2} & \\ & & v(x)=[w(x)]^{-2} & \\ & & v^{\prime}(x)=-2(2x^{3}+1)^{-3}\cdot w^{\prime}(x) & (\text{chain rule)})\\ & & v^{\prime}(z)=-2(2x^{3}+1)^{-3}\cdot 6x^{2} & \end{array}\right]$. $\displaystyle \frac{dy}{dx}=k\cdot v^{\prime}(x)$ $\displaystyle \frac{dy}{dx}=-5[-2(2x^{3}+1)^{-3}\cdot 6x^{2}]$ $\displaystyle \frac{dy}{dx}=-5[-12x^{2}(2x^{3}+1)^{-3}]$ $\displaystyle \frac{dy}{dx}=60x^{2}(2x^{3}+1)^{-3}$ $\displaystyle \frac{dy}{dx}=\frac{60x^{2}}{(2x^{3}+1)^{3}}$