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