Answer
$y'=\dfrac{\csc(x+1)^3}{2\sqrt x} -3\sqrt x \csc (x+1)^3 \cot (x+1)^3 (x+1)^2$
Work Step by Step
Given that $y=\sqrt x \csc(x+1)^3 $
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\dfrac{d}{dx}[\sqrt x \csc(x+1)^3]$
$=\dfrac{1}{2\sqrt x} \csc(x+1)^3+\sqrt x[-csc (x+1)^3 \cot (x+1)^3]\dfrac{d}{dx}(x+1)^3$
$=\dfrac{\csc(x+1)^3}{2\sqrt x} -\sqrt x \csc (x+1)^3 \cot (x+1)^3 \cdot 3(x+1)^2$
Hence, $y'=\dfrac{\csc(x+1)^3}{2\sqrt x} -3\sqrt x \csc (x+1)^3 \cot (x+1)^3 (x+1)^2$
