Then $x = \sin y$.

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

$1 = (\cos y){{dy} \over {dx}}$

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

or, ${{dy} \over {dx}} = {1 \over {\cos y}}$.

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

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