find $\frac{dy}{dt}$. $y = (\frac{t^{2}}{t^{3}-4t})^{3}$

Explanation:

Step1: Simplify the function

First, simplify $y = \left(\frac{t^{2}}{t^{3}-4t}\right)^{3}=\frac{t^{6}}{(t^{3}-4t)^{3}}$.

Step2: Apply the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = t^{6}$, so $u^\prime=6t^{5}$, and $v=(t^{3}-4t)^{3}$. To find $v^\prime$, use the chain - rule. Let $w=t^{3}-4t$, then $v = w^{3}$. By the chain - rule, $v^\prime = 3w^{2}\cdot w^\prime$. Since $w^\prime=3t^{2}-4$, then $v^\prime=3(t^{3}-4t)^{2}(3t^{2}-4)$.

Step3: Calculate $\frac{dy}{dt}$

[ \begin{align*} \frac{dy}{dt}&=\frac{6t^{5}(t^{3}-4t)^{3}-t^{6}\cdot3(t^{3}-4t)^{2}(3t^{2}-4)}{(t^{3}-4t)^{6}}\ &=\frac{(t^{3}-4t)^{2}[6t^{5}(t^{3}-4t)-3t^{6}(3t^{2}-4)]}{(t^{3}-4t)^{6}}\ &=\frac{6t^{8}-24t^{6}-9t^{8}+12t^{6}}{(t^{3}-4t)^{4}}\ &=\frac{- 3t^{8}-12t^{6}}{(t^{3}-4t)^{4}}\ &=\frac{-3t^{6}(t^{2} + 4)}{(t^{3}-4t)^{4}} \end{align*} ]

Answer:

$\frac{-3t^{6}(t^{2}+4)}{(t^{3}-4t)^{4}}$