Show that |x(t)|d/dt*|x(t)| = x(t)x`(t)

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Homework Statement



[itex]Show that |x(t)| d/dt |x(t)| = x(t) \cdot x`(t)[/itex]

The Attempt at a Solution



I don't know what they want to see here but my go at this is to take x(t) = 2t. Differentiating this gives x`(t) = 2.

Taking the normal of both |x(t)| and |x`(t)| gives 2t and 2, respectively.

So 2t * 2 = 4t.

Then just do the same with x(t) and x`(t) to get 4t again?

I don't know if this will suffice as a proper answer
 
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I assume they expect you to show the result for any path x(t), not just x(t)=2t .

I think working with the cases (for the absolute value) ; when is x(t)>0 , when is it <0 ,=0 ,

etc. will be helpful.
 
Hint: Use the fact that [itex]\frac{d}{dt} |t|=\frac{t}{|t|}[/itex] with the chain rule to compute your derivative in the two cases [itex]x(t) > 0[/itex] and [itex]x(t) < 0[/itex] . You have to consider the case [itex]x(t)=0[/itex] carefully.