Help with $\arcsin(x)$ derivative and differentials.

I'm watching this video lecture http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-11-chain-rule/ and I'm stuck at around 3:40, I can't seem to figure out what he is doing.

He is showing how to derive $f(x)=\sin^{-1}(x)$.

At some point he goes from the expression $\frac{dy}{dx}=\frac{1}{\cos(y)}$ to $\frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}$.

Ok, I know that's the result, that's how I always did it, but I never actually derived it myself. So yeah, I'd like to know what he did in those last two steps, to get from the first expression to the second.

I know it may (will) be something completely stupid and I'll say 'oh... facepalm', but for some reason I can't figure out how he did it. I'm guessing the next logical step is to replace $y=\sin^{-1}(x)$, but then? I've never been in good terms with trigonometric functions and identities really, so I'd appreciate some enlightening.

Thank you.

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Remember that (in a suitable interval) $\cos y = \sqrt{1-\sin^2 y}$, and that here $y = \arcsin x$, so $x = \sin y$ and $\cos y = \sqrt{1-x^2}$. $\frac{dy}{dx} = \frac1{\cos y}$ comes from the "formula" $\frac1{\frac{dx}{dy}} = \frac{dy}{dx}$.
Another way to deal with this, which is useful for obtaining all of the derivatives of the inverse trig functions, is to keep in mind that $\ y \ = \ \sin^{-1} x \ \Rightarrow \sin y \ = \ x \ = \frac{x}{1} \$. You can now construct a right triangle with one of the angles being $y$ : the leg opposite $y$ has length $x$, and the hypotenuse, a length of $1$ . The leg adjacent to $y$ must then have length $\sqrt{1 - x^2}$ . Hence,
$$\cos y \ = \ \frac{\sqrt{1 - x^2}}{1} \ \Rightarrow \frac{d}{dx}\sin^{-1} x \ = \ \frac{1}{\sqrt{1 - x^2}} .$$