Answer
$y'=\dfrac{-2 \cos x}{ \sin^3 x}+\dfrac{2 \cos x}{\sin^2 x}$
Work Step by Step
Given that $y=\dfrac{1}{ \sin^2 x}-\dfrac{2}{\sin x}$
Re-write the given equation as: $y= \sin^{-2} x-2\sin^{-1} x$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\frac{d}{dx}[ \sin^{-2} x-2\sin^{-1} x]$
$= (-2)\sin^{-3} x \cos x+2\sin^{-2} x \cos x$
Hence, $y'=\dfrac{-2 \cos x}{ \sin^3 x}+\dfrac{2 \cos x}{\sin^2 x}$
