We have to differentiate this function at $x=0$
$$\left. \dfrac{d \arcsin (\cos (x))}{dx} \right|_{x=0}$$
Using the identity $\cos x=\sin \Big(\dfrac{\pi}{2}-x\Big)$ we get
$$\left. \dfrac{d \arcsin (\sin (\dfrac{\pi}{2} – x))}{dx} \right|_{x=0},$$
For $0 \le x \le \pi$. We know
$\arcsin \Big(\sin \Big(\dfrac{\pi}{2}-x\Big)\Big)=\dfrac{\pi}{2}-x$.
So, our original problem reduces to
$$\left. \dfrac{d\Big(\dfrac{\pi}{2}-x\Big)}{dx}\right|_{x=0},$$
which is equal to $-1$ for $0 \le x \le \pi$.
But the derivative of this function at $x=0$ is undefined. What's going on here?
Best Answer
Near zero you have $$\arcsin(\cos x) = \frac{\pi}{2} - \arccos(\cos x) = \frac{\pi}{2} - |x|$$ (so the derivative is $+1$ for $x<0$ and $-1$ for $x>0$). And therefore the function cannot be approximated by a linear function in a arbitrary small interval containg $0,$ and this implies that the derivative at $x=0$ does not exist.
Edit: Near $x = 0$ you have $\arccos(\cos x)=|x|$ because $\cos(x)$ is an even function. It cannot be just $x$ because the value is positive for $x\ne0$.
You also have to look at values $x<0,$ if you want to know whether the function has a derivative in a neighborhood of $0$ (it has, if the left and rights derivatives exist and are equal, see e.g. here).