$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists. 
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.

My proof is here:

Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}}$.
So, $\lim_{h\to 0+} \frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = -\infty$.
So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.

I want an easier proof.
Is there any general theorem or general proposition to prove the above fact?
By the way, the following proposition is not true. $f(x) = x^{\frac{1}{3}}$ is not differentiable at $x = 0$ and $g(x) = x^3$ is differentiable at $x = 0$ but $f(g(x)) = x$ is differentiable at $x=0$.

If $f(x)$ is not differentiable at $x = a$ and $g(x)$ is differentiable at $x = b$ and $g(b) = a$, then $f(g(x))$ is not differentiable at $x = b$.

So, we cannot prove as follows:

$\sqrt{x}$ is not differentiable at $x = 0$. And $1-x^2$ is differentiable at $x = 1$. So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.


The derivative of $x\mapsto\sqrt{1-x^2}$ exists only for $-1<x<1$:
$$f'(x)=-\frac{x}{\sqrt{1-x^2}}$$
Since the function is continuous on $[-1,1]$, we can compute the limits of the derivative. Then
\begin{align}
\lim_{x\to1^+}f'(x)&=i\infty\\
\lim_{x\to 1^-}f'(x)&=-\infty
\end{align}
so the function is not differentiable at $x=1$ as we have a two-sided limit (see this also).
