Why in this derivation we have ${1\over{\cos y}}={1\over{(1-x^2)^{1/{2}}}}$?

Question

Let $\sin^{-1}x=y$, then $\sin y = x $ and therefore:

$$\eqalign{\cos y \,{dy\over{dx}} = 1&\Longrightarrow {dy\over{dx}} = {1\over{\cos y}}\\ &\Longrightarrow {dy\over{dx}} = {1\over{(1-x^2)^{1/{2}}}}}$$

Why in this derivation we have $\displaystyle{1\over{\cos y}}={1\over{(1-x^2)^{1/{2}}}}$?

Answer 1

Hint:
Use the Pythagorean identity $$\cos \;\!{y}=\pm\sqrt{1-\sin^2y},$$ what should be its sign based on the range of $y$ ? To finish it off use that $y=\sin^{-1}x$. Be careful with the domains and ranges!