Answer
$$\frac{dy}{dt}=3(2t^2-5)^3\Big(18t^2-5\Big)$$
Work Step by Step
$$\frac{dy}{dt}=\frac{d}{dt}\Big(3t(2t^2-5)^4\Big)$$
Following the Product Rule and Chain Rule: $$\frac{dy}{dt}=(2t^2-5)^4\frac{d}{dt}(3t)+(3t)\frac{d}{dt}(2t^2-5)^4$$
$$\frac{dy}{dt}=3(2t^2-5)^4+3t\times\Big(4(2t^2-5)^3\frac{d}{dt}(2t^2-5)\Big)$$
$$\frac{dy}{dt}=3(2t^2-5)^4+3t\times\Big(4(2t^2-5)^3(4t)\Big)$$
$$\frac{dy}{dt}=3(2t^2-5)^4+48t^2(2t^2-5)^3$$
$$\frac{dy}{dt}=(2t^2-5)^3\Big(3(2t^2-5)+48t^2\Big)$$
$$\frac{dy}{dt}=(2t^2-5)^3\Big(54t^2-15\Big)$$
$$\frac{dy}{dt}=3(2t^2-5)^3\Big(18t^2-5\Big)$$
