Check the continuity and differentiability of $f(x)= \sin^{-1}(\cos x)$ at $x=0$
This is how I tried to solve the problem:
$$f(x)= \sin^{-1}(\cos x)=???$$
$$\lim_{x \to 0^+}f(x)=\lim_{x \to 0^-}f(x)=f(0)=???$$
Therefore the function is continuous at $x=0$
$$\lim_{x \to 0^+}\frac{f(x)-f(0)}{x-0}=???$$and
$$\lim_{x \to 0^-}\frac{f(x)-f(0)}{-x}=???$$
Best Answer
In this question it's important to note that $\sin^{-1}(c)$ is (customarily) defined as the unique value of $x \in [-\pi/2,\pi/2]$ such that $\sin x = c$. In particular, $f(x) = \sin^{-1}(\cos x)$ can never take values outside the range $[-\pi/2,\pi/2]$, contradicting the claim in another answer that $f(x) = \pi/2 - x$.
If you graph the function carefully near $x=0$, you'll find it simplifies to $f(x) = \pi/2 - |x|$ for $-\pi \le x \le \pi$. I think you already know something about whether $|x|$ is differentiable at $x=0$.