#### Answer

$g'(θ) =-sin(2θ) $

#### Work Step by Step

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