find $\frac{dy}{dt}$. $y = (\frac{t^{2}}{t^{3}-6t})^{3}$ $\frac{dy}{dt}=square$

Explanation:

Step1: Apply the chain - rule

Let $u=\frac{t^{2}}{t^{3}-6t}$, so $y = u^{3}$. By the chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{dy}{du}$. Since $y = u^{3}$, then $\frac{dy}{du}=3u^{2}$.

Step2: Apply the quotient - rule to find $\frac{du}{dt}$

The quotient - rule states that if $u=\frac{f(t)}{g(t)}$ where $f(t)=t^{2}$ and $g(t)=t^{3}-6t$, then $\frac{du}{dt}=\frac{f^{\prime}(t)g(t)-f(t)g^{\prime}(t)}{g(t)^{2}}$. We know that $f^{\prime}(t) = 2t$ and $g^{\prime}(t)=3t^{2}-6$. So $\frac{du}{dt}=\frac{2t(t^{3}-6t)-t^{2}(3t^{2}-6)}{(t^{3}-6t)^{2}}=\frac{2t^{4}-12t^{2}-3t^{4}+6t^{2}}{(t^{3}-6t)^{2}}=\frac{-t^{4}-6t^{2}}{(t^{3}-6t)^{2}}$.

Step3: Substitute $u$ and $\frac{du}{dt}$ into the chain - rule formula

$\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}=3u^{2}\cdot\frac{-t^{4}-6t^{2}}{(t^{3}-6t)^{2}}$. Substitute $u = \frac{t^{2}}{t^{3}-6t}$ back in: $\frac{dy}{dt}=3\left(\frac{t^{2}}{t^{3}-6t}\right)^{2}\cdot\frac{-t^{4}-6t^{2}}{(t^{3}-6t)^{2}}=\frac{3t^{4}(-t^{4}-6t^{2})}{(t^{3}-6t)^{4}}=\frac{-3t^{8}-18t^{6}}{(t^{3}-6t)^{4}}$.

Answer:

$\frac{-3t^{8}-18t^{6}}{(t^{3}-6t)^{4}}$