$f$ is differentiable when $f^2$ is differentiable

Question

My textbook claims that for a continuous function $f$, if $f^2$ is differentiable, then $f$ is differentiable at x whenever $f(x) \neq 0$. The claim is given without proof. I also found the following question on math stack exchange that makes the same claim.

Can someone help me by pointing to a proof (or give a proof)?

You can prove the reverse. i.e. if $f$ is continuous at $x_0$ and not differentiable there, then $f^2$ is not differentiable either — polfosol, Oct 31 '16 at 07:55
Alternatively, note that both $f^2(x)$ and $g(x) = \sqrt{x}$ are differentiable on appropriate domains, so their composition is too. — Guy Paterson-Jones, Oct 31 '16 at 07:56
@polfosol That is called the contrapositive, not the reverse. Still, it's a very valid tactic. — Arthur, Oct 31 '16 at 07:59

score 5 · Accepted Answer · edited Oct 31 '16 at 08:09

5

Suppose $f(x_0) \ne 0$. Then $f(x) \ne 0$ for $x$ near by $x_0$, since f is continuous. Now we have

$$\frac{f(x)-f(x_0)}{x-x_0}=\frac{f^2(x)-f^2(x_0)}{x-x_0}\frac{1}{f(x)+f(x_0)}$$

for $x$ near by $x_0$.

edited Oct 31 '16 at 08:09

answered Oct 31 '16 at 08:05

Fred

77,394

score 1 · Answer 2 · answered Oct 31 '16 at 08:14

$$\displaystyle\frac{f(a+h)-f(a)}h=\frac{f^2(a+h)-f^2(a)}{h(f(a+h)+f(a))}$$

$$\displaystyle ⇒\frac{(f^2(a))'}{2f(a)}$$

by $\displaystyle (f^2(a))'=2f(a)f'(a)$

Then $$\displaystyle \frac{f(a+h)-f(a)}h=f'(a)$$

So If $f^2(x)$ is differentiable, $f(x)$ is differentiable, too.

$f$ is differentiable when $f^2$ is differentiable

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