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

Question

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 $x = 1$.

Answer 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).

Answer 2

$$\lim_{x\to1}\frac{\sqrt{1-x^2}-0}{x-1}=\lim_{x\to1}\sqrt{\frac{1+x}{1-x}}.$$

You can conclude.

Answer 3

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.

Answer 4

Here's another way . . .

On the domain $[-1,1]$, let $f(x)=\sqrt{1-x^2}$ and let $g=f^2$.

If $f$ was differentiable at $x=1$ then by the power rule, the equation $$ g'=2ff' $$ would hold at $x=1$, but at $x=1$ the $\text{LHS}$ evaluates to $-2$ (since $g'(x)=-2x$) while the $\text{RHS}$ evaluates to $0$ (since $f(1)=0$).

Answer 5

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.