## Thomas' Calculus 13th Edition

$\frac{3t^2(t^2+4)}{(t^2-4)^4}$
y=$(\frac{t^2}{t^3-4t})^3$ Apply the chain rule: $\frac{dy}{dt}=3(\frac{t^2}{t^3-4t})^2.\frac{(t^3-4t)(2t)-t^2(3t^2-4).5}{(t^3-4t)^2}$ =$\frac{3t^4}{(t^3-4t)^2}.\frac{2t^4-8t^2-3t^4+4t^2}{(t^3-4t)^2}$ =$\frac{3t^4(-t^4-4t^2)}{t^4(t^2-4t)^4}$ =$\frac{3t^2(t^2+4)}{(t^2-4)^4}$