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