$\sqrt{1-x^2}$ 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 $x = 1$.


Let $g(x) = 1 - x^2$. Then
\begin{align}
g(f(x)) 
&= 1 - \left(\sqrt{1-x^2} \right)^2\\
&= 1 - \left(1-x^2 \right)\\
&= x^2\\
\end{align}
so
$$
(g \circ f)'(x) = 2x \tag{1}
$$ for every $x$.
$f$ is evidently differentiable for $-1 < x < 1$.
Suppose that $f$ were differentiable at $x = 1$ as well.
Then (by a version of the chain rule extended to deal with one-sided limits), we'd have
$$
(g\circ f)'(1) = g'(f(1)) \cdot f'(1)
$$
Using equation $1$, this becomes
$$
2 = g'(0) \cdot f'(1)
$$
Now $g'(x) = -2x$, so $g'(0) = 0$, so we would have (assuming $f$ is differentiable!) that
$$
2 = 0 \cdot f'(1)
$$
which is impossible. Hence our assumption that $f$ is differentiable (at $1$) must be incorrect.
Is this "simpler"? Probably not. After all, you have to review your proof of the chain rule to confirm that it works when some/all limits are one-sided. That's straightforward, but involves some writing.
