## University Calculus: Early Transcendentals (3rd Edition)

$$\frac{dy}{dt}=\frac{3t^2(t^2-4)-6t^4}{(t^2-4)^4}$$
$$\frac{dy}{dt}=\frac{d}{dt}\Big(\frac{t^2}{t^3-4t}\Big)^3=\frac{d}{dt}\Big(\frac{t}{t^2-4}\Big)^3=\frac{d}{dt}\frac{t^3}{(t^2-4)^3}$$ According to the Quotient Law: $$\frac{dy}{dt}=\frac{(t^2-4)^3\frac{d}{dt}(t^3)-t^3\frac{d}{dt}(t^2-4)^3}{(t^2-4)^6}$$ $$\frac{dy}{dt}=\frac{(t^2-4)^3\times(3t^2)-t^3\Big(3(t^2-4)^2\frac{d}{dt}(t^2-4)\Big)}{(t^2-4)^6}$$ $$\frac{dy}{dt}=\frac{3t^2(t^2-4)^3-3t^3(t^2-4)^2(2t)}{(t^2-4)^6}$$ $$\frac{dy}{dt}=\frac{3t^2(t^2-4)^3-6t^4(t^2-4)^2}{(t^2-4)^6}$$ $$\frac{dy}{dt}=\frac{3t^2(t^2-4)-6t^4}{(t^2-4)^4}$$