Here's one way of looking at it: Let $y = \arcsin x$ so $x = \sin y$. Then $\frac{dx}{dy} = \cos y$, so $\frac{dy}{dx} = \frac{1}{\cos y}$. Since $\sin^2 y + \cos^2 y = 1$, you can change $\frac{1}{\cos y}$ to $\frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1-x^2}}$.
Two questions arise: why is there no "$\pm$" in front of the radical? and is the reciprocal of $\frac{dx}{dy}$ really $\frac{dy}{dx}$ even though $dy$ and $dx$ aren't actually numbers?
The answer to the first question comes from the fact that if $y = \arcsin x$ then $y$ is between $-\pi/2$ and $\pi/2$, so that $\cos y$ is nonnegative (either positive or zero).
The answer to the second question is "yes" because of the chain rule.