Proving $\lim_{h \to 0} \frac {\arccos(\cos^2h)} {h}$ doesn't exist I want to show $\arcsin (\sin^2x) $ is not differentiable at $\pi/2+\pi k$. (if its true).
So far I have:
$$ \frac{\arcsin (\sin^2(\pi/2+\pi k +h))-\arcsin (\sin^2(\pi/2+\pi k))}{h}=\frac{\arcsin (\cos^2h)-\ \pi/2}{h} = \frac {\arccos (\cos^2 h)}{h} $$
so it's sufficient to wish to show this limit exists or doesn't exist:
$$\lim_{h \to 0} \frac {\arccos(\cos^2h)} {h}$$
Not sure how to proceed. 
EDIT: I edited the question to cover $-\pi/2 +2\pi k$ aswell. I feel the question didn't get enough attention.
I have been told by a friend it's solveable using l'hopitals rule looking at the one sided limits, however l'hopitals is 2 months ahead in my course. a little strange this problem appears now.
 A: Since the function has period $\pi$, computing the limit at $\pi/2$ is sufficient. Now, where's the error?
The computation
$$
\arcsin\left(\sin^2\left(\frac{\pi}{2}+h\right)\right)-\arcsin\sin^2\frac{\pi}{2}=
\arcsin\cos^2 h-\frac{\pi}{2}=-\arccos\cos^2h
$$
is correct. Now your limit is exactly computing the derivative of $f(x)=\arccos\cos^2x$ at $0$.
The function is continuous and its derivative, except possibly at zero, exists:
$$
f'(x)=\frac{2\sin x\cos x}{\sqrt{1-\cos^2x}}=
\frac{2\sin x\cos x}{|\sin x|}
$$
and the limit of this at $0$ is different when computed from the left and from the right:
$$
\lim_{x\to 0+}f'(x)=1,\quad
\lim_{x\to 0-}f'(x)=-1.
$$
Now, I know that this uses l'Hôpital's theorem, but this is a guide. If $f$ were differentiable at $0$, the derivative should be $0$, because the function is even and non negative. So we want to show that
$$
\frac{f(x)}{x}\ge1
$$
holds for $0<x<k$ (for some $k>0$). This means
$$
\arccos\cos^2x>x
$$
or, since the cosine is decreasing in $[0,\pi/2]$,
$$
\cos^2x<\cos x
$$
which is true for $0<x<\pi/2$.
