Answer
$r'=2 \sqrt { \cos \theta}-\dfrac{ \theta \sin \theta}{\sqrt { \cos \theta}}$
Work Step by Step
Given that $r=2 \theta\sqrt { \cos \theta}$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$r'=\dfrac{d}{dx}[2 \theta\sqrt { \cos \theta}]$
$=2 \sqrt { \cos \theta}+\dfrac{2 \theta}{2\sqrt { \cos \theta}}\dfrac{d}{dx}( \cos \theta)$
$=2 \sqrt { \cos \theta}-\dfrac{2 \theta \sin \theta}{2\sqrt { \cos \theta}}$
Hence, $r'=2 \sqrt { \cos \theta}-\dfrac{ \theta \sin \theta}{\sqrt { \cos \theta}}$
