Proving derivatives are equal

The question is:

Assume $f \colon\mathbb{R}\to\mathbb{R}$, $c\in \mathbb{R}$, $f(c)=0$, $f$ is differentiable at $c$, and $g\colon\mathbb{R}\to\mathbb{R}$ is defined by $g(x)=|f(x)|$. Show that if $f'(c)=0$, then $g$ is differentiable at $c$ and $g'(c)=0$.

So what I did was say that $$g'(c) = \lim_{x\to c} \frac{|f'(x)|-|f'(c)|}{x-c} \leq \left|\lim_{x\to c} \frac{f'(x)-f'(c)}{x-c}\right| = |f'(c)|$$ so $g'(c) \leq |f'(c)|$; but after this I am stuck. I am trying to use squeeze theorem, but am not sure if I am on the right track.

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The inequality is not necessary strict. – Arturo Magidin Nov 17 '11 at 17:00
Thanks I just changed it – rioneye Nov 17 '11 at 17:30
The conclusion also needed changing. In any case, I've typeset your question and made that correction. – Arturo Magidin Nov 17 '11 at 17:42

$$g'(c)= \lim_{x \to c} \frac { |f(x)|}{x-c}$$
Squeeze as , $$\frac { -f(x)}{x-c} \le \frac { |f(x)|}{x-c} \le \frac { f(x)}{x-c} .$$ Apply L'Hopital Rule to obtain the limits $x \to c$. You are done.