Answer
$$\frac{dy}{dt}=-t\sin t^2(1+\cos(t^2))^{-1/2}$$
Work Step by Step
$$\frac{dy}{dt}=\frac{d}{dt}\sqrt{1+\cos(t^2)}=\frac{d}{dt}(1+\cos(t^2))^{1/2}$$
Following the Chain Rule: $$\frac{dy}{dt}=\frac{1}{2}(1+\cos(t^2))^{-1/2}\frac{d}{dt}(1+\cos(t^2))$$
$$\frac{dy}{dt}=\frac{1}{2}(1+\cos(t^2))^{-1/2}(0-\sin t^2\frac{d}{dt}(t^2))$$
$$\frac{dy}{dt}=\frac{1}{2}(1+\cos(t^2))^{-1/2}(-2t\sin t^2)$$
$$\frac{dy}{dt}=-t\sin t^2(1+\cos(t^2))^{-1/2}$$
