Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 50

$\frac{dy}{dt}$ = $[-\frac{5}{3}cos(\frac{t}{3})]$$sin(5sin(\frac{t}{3}))$

$y$ = $cos(5sin(\frac{t}{3}))$ $\frac{dy}{dt}$ = $\frac{d}{dt}$$cos(5sin(\frac{t}{3}))$ $\frac{dy}{dt}$ = -$sin(5sin(\frac{t}{3}))$ $\frac{d}{dt}$$(5sin(\frac{t}{3}))$ $\frac{dy}{dt}$ = -$sin(5sin(\frac{t}{3}))$$[5cos(\frac{t}{3})]$ $\frac{d}{dt}$$(\frac{t}{3})$ $\frac{dy}{dt}$ = -$sin(5sin(\frac{t}{3}))$$[5cos(\frac{t}{3})]$$(\frac{1}{3})$ $\frac{dy}{dt}$ = $[-\frac{5}{3}cos(\frac{t}{3})]$$sin(5sin(\frac{t}{3}))$