Let $y = f(x) = \csc ^{ - 1} x$

Then $x = \csc y$.

Differentiating both sides with respect to $x$, we get

$1 = (-\csc y\cot y){{dy} \over {dx}}$

Note that we used the chain rule to differentiate $\csc y$ with respect to $x$ because $\csc y$ is a function of $y$, and $y$ in turn is a function of $x$.

or, ${{dy} \over {dx}} = {-1 \over {\csc y\cot y}}$

or, ${{dy} \over {dx}} = {-1 \over {\csc y\sqrt {\csc ^2 y - 1} }}$

or, ${{dy} \over {dx}} = {-1 \over {x\sqrt {x^2 - 1} }}$