Answer
$s'=\dfrac{( \sin \theta + \theta \cos \theta)}{ \sqrt{2\theta \sin \theta}}$
Work Step by Step
Given that $r=\sqrt {2 \theta \sin \theta}$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$r'=\dfrac{d}{dx}[\sqrt {2 \theta \sin \theta}]$
$=(2 \sqrt{2\theta \sin \theta})^{-1}\dfrac{d}{dx}[ {2 \theta \sin \theta}]$
$=(2 \sqrt{2\theta \sin \theta})^{-1}( 2 \sin \theta +2 \theta \cos \theta)$
Hence, $s'=\dfrac{( \sin \theta + \theta \cos \theta)}{ \sqrt{2\theta \sin \theta}}$
