Answer
$y'=-4\sqrt{3}$.
Work Step by Step
Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=1+\csc(x); u'(x)=-\csc(x)\cot(x)$
$v(x)=1-\csc(x); v'(x)=\csc(x)\cot(x)$
$y'=\frac{-\csc(x)\cot(x)-\csc(x)\cot(x)}{(1-\csc(x))^2}$
$=-\frac{2\csc(x)\cot(x)}{(1-\csc(x))^2}$.
To evaluate, substitute in $x=\frac{\pi}{6}$:
$y'=-\dfrac{2\csc(\frac{\pi}{6})\cot(\frac{\pi}{6})}{(1-\csc(\frac{\pi}{6}))^2}=-4\sqrt{3}$
