Answer
$\displaystyle \frac{dy}{dx}=\frac{60x^{2}}{(2x^{3}+1)^{3}}$
Work Step by Step
$y=\displaystyle \frac{-5}{(2x^{3}+1)^{2}}=-5(2x^{3}+1)^{-2}=\mathrm{k}\cdot v(x)$
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.
$\left[\begin{array}{llll}
& & v(x)=(2x^{3}+1)^{-2} & \\
& & v(x)=[w(x)]^{-2} & \\
& & v^{\prime}(x)=-2(2x^{3}+1)^{-3}\cdot w^{\prime}(x) & (\text{chain rule)})\\
& & v^{\prime}(z)=-2(2x^{3}+1)^{-3}\cdot 6x^{2} &
\end{array}\right]$.
$\displaystyle \frac{dy}{dx}=k\cdot v^{\prime}(x)$
$\displaystyle \frac{dy}{dx}=-5[-2(2x^{3}+1)^{-3}\cdot 6x^{2}]$
$\displaystyle \frac{dy}{dx}=-5[-12x^{2}(2x^{3}+1)^{-3}]$
$\displaystyle \frac{dy}{dx}=60x^{2}(2x^{3}+1)^{-3}$
$\displaystyle \frac{dy}{dx}=\frac{60x^{2}}{(2x^{3}+1)^{3}}$
