Where does this proof for arccosine derivative break down? So I'm checking the identities for the derivative of the inverse cosine, and I ran against a wall:
With $\cos(y)=x$  
$dy\over dx$$=$$1\over\sin(-y)$$=$$1\over-\sin(y)$  
From here, I could either do $1\over\sqrt{1-\cos^2(-y)}$$=$$1\over\sqrt{1-\cos^2(y)}$  $=$$1\over\sqrt{1-x^2}$
Since cosine is an even function.   
Or I could go the simple way in which I get the correct answer:
$-1\over\sqrt{1-\cos^2(y)}$$=$$-1\over\sqrt{1-x^2}$
Why is my first answer incorrect? I suspect it was when I used the pythagorean identity, but I can't see why I'd get two different answers if I'm using identities only.
Thanks in advance.

The inverse cosine has domain $[-1,1]$ and range $[0,\pi]$, so your $y$ is always between $0$ and $\pi$. This means that $\sin y\ge 0$ and $\sin(-y)\le 0$. Thus, $\sin(-1)$ must be the negative square root of $1-\cos^2(-y)$, i.e.,
$$\sin(-y)=-\sqrt{1-\cos^2(-y)}=-\sqrt{1-\cos^2y}\;.$$
Now the first method also gives you the correct result.
