Answer
$$\frac{dy}{dt}=-\pi e^{\cos^2(\pi t-1)}\sin(2\pi t-2)$$
Work Step by Step
$$\frac{dy}{dt}=\frac{d}{dt}e^{\cos^2(\pi t-1)}$$
According to the Chain Rule:
$$\frac{dy}{dt}=e^{\cos^2(\pi t-1)}\frac{d}{dt}\Big(\cos^2(\pi t-1)\Big)'$$
$$\frac{dy}{dt}=e^{\cos^2(\pi t-1)}2\cos(\pi t-1)\frac{d}{dt}(\cos(\pi t-1))$$
$$\frac{dy}{dt}=e^{\cos^2(\pi t-1)}2\cos(\pi t-1)(-\sin(\pi t-1))\frac{d}{dt}(\pi t-1)$$
$$\frac{dy}{dt}=-e^{\cos^2(\pi t-1)}2\cos(\pi t-1)\sin(\pi t-1)(\pi-0)$$
$$\frac{dy}{dt}=-\pi e^{\cos^2(\pi t-1)}2\cos(\pi t-1)\sin(\pi t-1)$$
Recall the identity $2\cos\alpha\sin\alpha=\sin2\alpha$:
$$\frac{dy}{dt}=-\pi e^{\cos^2(\pi t-1)}\sin(2\pi t-2)$$