<s>[INST] $\sqrt{1-x^2}$ 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 $x = 1$.

 [/INST] It all depends on how much you "want to prove".
The function $f(x)= \sqrt{1-x^2}$ has domain $x\in [-1,1]$. The derivative of $f$, if well-defined, can only be defined on $(-1,1)$. If possible, you can extend the definition to $x=\pm 1$ by taking the left- and right-limits.
The derivative of $f$ is
$$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{\sqrt{1-(x+h)^2} - \sqrt{1-x^2}}{h} = \lim_{h\to 0} \frac{\sqrt{1-(x+h)^2} - \sqrt{1-x^2}}{h}\frac{\sqrt{1-(x+h)^2} + \sqrt{1-x^2}}{\sqrt{1-(x+h)^2} + \sqrt{1-x^2}}= \lim_{h\to 0} \frac{1-(x+h)^2 -1+x^2}{\sqrt{1-(x+h)^2} + \sqrt{1-x^2}}=\lim_{h\to 0} \frac{-2x-h}{\sqrt{1-(x+h)^2} + \sqrt{1-x^2}} = -\frac{x}{\sqrt{1-x^2}}.$$
As you can see the function $f$ is not defined on $x=-1$ or $x=1$, hence $f$ is not differentiable at those points. There is not much proof other than differentiating $f$ and looking at its definition domain.
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