The derivative of $\arccos(x)$ and $\arcsin(x)$ are respectively $-\frac 1{\sqrt{1-x^2}}$ and $\frac 1{\sqrt{1-x^2}}$. However in the classic proof using implicit differentiation, when taking the square root, from $\sin^2y + \cos^2y = 1$, we should get ± $\sqrt{1-\sin^2(\arcsin(x))}$ and ± $\sqrt{1-\cos^2(\arccos(x))}$ respectively. So, what is the difference between the two derivatives? If we only consider the positive root, why do we do that?
Edit: i am refering to $\operatorname{arccos}'x=\frac1{\cos'(\arccos{x})}=-\frac1{\sin\arccos{x}}$ as pointed out by Giulio
It can be shown that $\forall x \in [-1, 1]$:
$$ \arcsin(x) + \arccos(x) = \frac{\pi}{2}$$
Implicit differentiation gives:
$$ \arcsin'(x) + \arccos'(x) = 0$$ $$ \arccos'(x) = -\arcsin'(x)$$
Thus, the derivatives of the two functions must be same except for having opposite signs.
That $\arcsin'$ is positive and $\arccos'$ is negative follows from $\arcsin$ being increasing and $\arccos$ being decreasing.