Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 43

$\frac{dy}{dt}$ = $\frac{8sin(2t)}{(1+cos(2t))^{5}}$

$y$ = $(1+cos(2t))^{-4}$ $\frac{dy}{dt}$ = $\frac{d}{dt}$$(1+cos(2t))^{-4}$ $\frac{dy}{dt}$ = $(-4)(1+cos(2t))^{-5}$ $\frac{d}{dt}$$(1+cos(2t))$ $\frac{dy}{dt}$ = $(-4)(1+cos(2t))^{-5}$$(-sin(2t))$ $\frac{d}{dt}$$(2t)$ $\frac{dy}{dt}$ = $(-4)(1+cos(2t))^{-5}$$(-sin(2t))$ $(2)$ $\frac{dy}{dt}$ = $(8sin(2t))(1+cos(2t))^{-5}$ $\frac{dy}{dt}$ = $\frac{8sin(2t)}{(1+cos(2t))^{5}}$