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