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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

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    $\begingroup$ Since $\arccos$ is decreasing, we have to take the negative square root. And, since $\arcsin$ is increasing, we have to take the positive square root. $\endgroup$ Commented Oct 5, 2023 at 21:37
  • $\begingroup$ $\arccos x\in[0,\pi]$ hence $\sin(\arccos x))\ge0.$ Similarly, $\arcsin x\in[-\pi/2,\pi/2]$ hence $\cos(\arcsin x))\ge0.$ As for the minus sign on the derivative of $\arccos,$ it is due to the minus sign in $\cos'=-\sin.$ $\endgroup$ Commented Oct 5, 2023 at 21:38

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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.

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