Derivative of Cosecant

Let $y = f(x) = \csc x$.

Since $\csc x = {1 \over {\sin x}} = (\sin x)^{ - 1} $, we use the chain rule and get

${{dy} \over {dx}} = ( - 1)(\sin x)^{ - 2} (\cos x)$

$ = {{ - 1} \over {\sin ^2 x}}\cos x$

$ = {{ - 1} \over {\sin x}}.{{\cos x} \over {\sin x}}$

$ =  - \csc x\cot x$