## University Calculus: Early Transcendentals (3rd Edition)

$$\frac{dy}{dt}=-7\sin7t\cos7t(1+\cos^2(7t))^2$$
$$\frac{dy}{dt}=\frac{d}{dt}\frac{1}{6}\Big(1+\cos^2(7t)\Big)^3$$ Following the Chain Rule: $$\frac{dy}{dt}=\frac{1}{6}\times3\Big(1+\cos^2(7t)\Big)^2\frac{d}{dt}(1+\cos^2(7t))$$ $$\frac{dy}{dt}=\frac{1}{2}(1+\cos^2(7t))^2(0+2\cos7t\frac{d}{dt}(\cos7t))$$ $$\frac{dy}{dt}=\frac{1}{2}(1+\cos^2(7t))^2(2\cos7t(-\sin7t)\frac{d}{dt}(7t))$$ $$\frac{dy}{dt}=\frac{1}{2}(1+\cos^2(7t))^2(-14\cos7t\sin7t)$$ $$\frac{dy}{dt}=-7\sin7t\cos7t(1+\cos^2(7t))^2$$