# Is there any function in this way?

$f$ is a function which is continous on $\Bbb R$, and $f^2$ is differentiable at $x=0$. Suppose $f(0)=1$. Must $f$ be differentiable at $0$?

I may feel it is not necessarily for $f$ to be differentiable at $x=0$ though $f^2$ is. But I cannot find a counterexample to disprove this. Anyone has an example?

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One note: some people may thing $\,f^2\,$ means the composition of $\,f\,$ with itself, whereas I think you actually meant the function squared...If I'm correct, then perhaps it'd be better to denote that as $\,f(x)^2\,$. – DonAntonio Dec 12 '12 at 3:08

Differentiability of $f$ is existence of $\lim(f(x)-1)/x$. Differentiability of $f^2$ is existence of $\lim((f(x))^2-1)/x$. Consider factoring the numerator of the latter and thinking about the consequences.

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SO the conclusion will be f must be differentiable at 0 if f^2 is differentiable at 0? – Scorpio19891119 Dec 12 '12 at 3:19
Sure looks that way. – Gerry Myerson Dec 12 '12 at 3:33

Hint:

$$\frac{f(x)-1}{x}=\frac{f(x)^2-1}{x}\frac{1}{f(x)+1}\xrightarrow [x\to 0]{}...?$$

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So... the conclusion will be f must be differentiable at 0? – Scorpio19891119 Dec 12 '12 at 3:18
Well...does the limit hinted at above exist finitely? I'd say it does... – DonAntonio Dec 12 '12 at 3:43