Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 191: 34

$y' = -4cos(4cos (z))sin(z)$

$y = sin (4cos (z) )$ $y' = (sin (4cos (z) ))'$ The inner function is $g(z) = 4cos (z)$ and the outer function is $f(u) = sin(u)$ $f'(u) = cos(u)$ $y' = (sin (4cos (z)))' = cos(g(z))g'(z) = cos(4cos (z))\times -4sin(z) = -4cos(4cos (z))sin(z)$