Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 41

$\frac{dy}{dt}$ = $(2\pi)$ $sin[\pi(t)-2]$ $cos[\pi(t)-2]$

$y$ = $sin^{2}[\pi(t)-2]$ $\frac{dy}{dt}$ = $\frac{d}{dt}$ [$sin^{2}[\pi(t)-2]$] $\frac{dy}{dt}$ = $(2)sin[\pi(t)-2]$ $\frac{d}{dt}$$sin[\pi(t)-2]$ $\frac{dy}{dt}$ = $(2)sin[\pi(t)-2]$ $cos[\pi(t)-2]$ $\frac{d}{dt}$ $[\pi(t)-2]$ $\frac{dy}{dt}$ = $(2)sin[\pi(t)-2]$ $cos[\pi(t)-2]$ ($\pi$) $\frac{dy}{dt}$ = $(2\pi)$ $sin[\pi(t)-2]$ $cos[\pi(t)-2]$