## Thomas' Calculus 13th Edition

-7$[1+cos^2(7t)]^2(cos(7t)sin(7t))$
y=$\frac{1}{6}[1+cos^2(7t)]^3$ Apply the chain rule: $\frac{dy}{dt}=\frac{3}{6}[1+cos^2(7t)]^2.2cos(7t)(-sin(7t)(7))$ =-7$[1+cos^2(7t)]^2(cos(7t)sin(7t))$