Let $y = f(x) = \cos ^{ - 1} x$
Then $x = \cos y$.
Differentiating both sides with respect to $x$, we get
$1 = ( - \sin y){{dy} \over {dx}}$
Note that we used the chain rule to differentiate $\cos y$ with respect to $x$ because $\cos y$ is a function of $y$, and $y$ in turn is a function of $x$.
or, ${{dy} \over {dx}} = {-1 \over {\sin y}}$.
or, ${{dy} \over {dx}} = {-1 \over {\sqrt {1 - \cos ^2 y} }}$
or, ${{dy} \over {dx}} = {-1 \over {\sqrt {1 - x^2 } }}$
