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