Answer
$$\frac{dy}{dt}=\tan^3\Big(\frac{t}{12}\Big)\sec^2\Big(\frac{t}{12}\Big)\Big(1+\tan^4\Big(\frac{t}{12}\Big)\Big)^2$$
Work Step by Step
$$\frac{dy}{dt}=\frac{d}{dt}\Big(1+\tan^4\Big(\frac{t}{12}\Big)\Big)^3$$
Following the Chain Rule: $$\frac{dy}{dt}=3\Big(1+\tan^4\Big(\frac{t}{12}\Big)\Big)^2\frac{d}{dt}\Big(1+\tan^4\Big(\frac{t}{12}\Big)\Big)$$
$$\frac{dy}{dt}=3\Big(1+\tan^4\Big(\frac{t}{12}\Big)\Big)^2\Big(0+4\tan^3\Big(\frac{t}{12}\Big)\frac{d}{dt}\Big(\tan\frac{t}{12}\Big)\Big)$$
$$\frac{dy}{dt}=3\Big(1+\tan^4\Big(\frac{t}{12}\Big)\Big)^2\Big(4\tan^3\Big(\frac{t}{12}\Big)\sec^2\frac{t}{12}\frac{d}{dt}\Big(\frac{t}{12}\Big)\Big)$$
$$\frac{dy}{dt}=3\Big(1+\tan^4\Big(\frac{t}{12}\Big)\Big)^2\Big(4\tan^3\Big(\frac{t}{12}\Big)\sec^2\frac{t}{12}\times\Big(\frac{1}{12}\Big)\Big)$$
$$\frac{dy}{dt}=\tan^3\Big(\frac{t}{12}\Big)\sec^2\Big(\frac{t}{12}\Big)\Big(1+\tan^4\Big(\frac{t}{12}\Big)\Big)^2$$
