Okay so this is my attempt of proving a, this was probably one of the most confusing things I did in my 2 month career in calculus with all the $x$'s and $y$'s and figuring out which is which so I'm not confident at all about the solution…
Regarding b, I just couldn't figure it out with the tool we currently have… we cannot use L'Hopital and we've just only started messing around with differential functions, so I'm stuck with picking the right variable to switch to in the one sided limits…
Question 1.
a.
Consider the function $g: (-1,1) -> \Bbb R, g(x) = \cos(x)$.
Note that we've seen in class that $g$ is both continuous and differential on $(-1,1)$.
We also know that $g'(x) = -\sin(x) \ne 0$, and $g^{-1}(y) = \arccos(y)$ for every $y \in (-1,1)$.
Also, by the algebra of continuous functions $g^{-1}$ is continuous in $(-1,1)$.
Therefore, we have
$f'(y) =\left(g^{-1}\right)^{'}\left(y\right)=\frac{1}{g'\left(x\right)}=\frac{1}{-\sin\left(x\right)}=\frac{1}{-\sin\left(\arccos\left(y\right)\right)}$
Note that we know $\sin^{2}\left(\arccos y\right)+\cos^{2}\left(\arccos y\right)=1$, then we have
$\sin^{2}\left(\arccos y\right)=1-\cos^{2}\left(\arccos y\right)$
$\sin^{2}\left(\arccos y\right)=1-y^{2}$
$\sin\left(\arccos y\right)=\pm\sqrt{1-y^{2}}$
Note that for every $y \in (-1,1)$, $\sin\left(\arccos y\right) > 0$, Therefore
$\sin\left(\arccos y\right)=\sqrt{1-y^{2}}$, which means
$f'(x) = \frac{1}{-\sin\left(\arccos x\right)}=\frac{-1}{\sqrt{1-x^{2}}}$
b.
Assume toward contradiction that $f$ is differentiable at $1$. Then we have
$\lim _{x\to 1^-}\frac{\left(\arccos\left(x\right)-\arccos\left(1\right)\right)}{x-1}=\lim_{x\to \:1^-}\frac{\left(\arccos\left(x\right)-0\right)}{x-1}$
We use a change of variable, $t = \arccos x$.
$\lim _{t\to 0^+}\frac{\left(t\right)}{\cos\left(t\right)-1}=\lim _{t\to 0^+}\frac{\left(t\right)}{\cos\left(t\right)-1}\cdot \frac{t}{t}=\lim _{t\to 0^+}\frac{\left(t\right)^2}{\cos\left(t\right)-1}\cdot \frac{1}{t}$
$=\lim _{t\to 0^+}\frac{1}{\frac{\left(\cos\left(t\right)-1\right)}{t^2}}\cdot \frac{1}{t}=\lim_{t\to0^+}-2\cdot \frac{1}{0^+}=-\infty$
Since the limit doesn't exist, we have a contradiction.
Assume toward contradiction f is differentiable at $-1$. Then we have
$\lim _{x\to -1^+}\frac{\left(\arccos\left(x\right)-\arccos\left(-1\right)\right)}{x+1}=\lim \:_{x\to \:-1^+}\frac{\left(\arccos\left(x\right)-\pi \right)}{x+1}$
We use a change of variable, $t = \arccos x$.
$\lim _{t\to \pi^+}\frac{\left(t\right)-\pi }{\cos\left(t\right)+1}$
We use a change of variable, $u = t – \pi$.
Note that $\cos(x + \pi) = -\cos(x)$.
$\lim _{u\to 0^-}\left(\frac{u}{cos\left(u+\pi \right)+1}\right)=\lim _{u\to 0^-}\left(\frac{u}{-cos\left(u\:\right)+1}\right)\cdot \frac{\left(cos\left(u\right)\:+1\right)}{cos\left(u\right)+1}=\lim _{u\to 0^-}\left(\frac{u\cdot \left(cos\left(u\right)+1\right)}{sin^2\left(u\right)}\right)=\lim _{u\to 0^-}\frac{\left(u\right)}{sin\left(u\right)}\cdot \frac{\left(u\right)}{sin\left(u\right)}\cdot \frac{1}{u}\cdot \left(cos\left(u\right)+1\right)=-\infty $
Since the limit doesn't exist, we have a contradiction.
Best Answer
Suppose that $\arccos$ is differentiable at $1$. The $\cos\circ\arccos$ is also differentiable at that point. But\begin{align}(\cos\circ\arccos)'(1)&=\cos'(\arccos(1))\arccos'(1)\\&=-\sin(0)\arccos'(1)\\&=0.\end{align}However, $\left(\forall x\in[-1,1]\right):\cos\bigl(\arccos(x)\bigr)=x$, and therefore $(\cos\circ\arccos)'(1)=1$. The same argument works for $-1$