Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises - Page 158: 14

$g'(θ) =-sin(2θ) $

Original Expression: $g(x)= cos^2(θ)$ $u = cos(θ)$ $g(u) = u^2$ Apply the chain rule: $g'(u)u'$ $g'(u) = 2(u)(-sin(θ))$ $g'(θ) = 2cos(θ)(-sin(θ))$ $g'(θ) =-2cos(θ)sin(θ) $ $g'(θ) =-sin(2θ) $, based on the double angle formula