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I know this is a stupid question, but it has been a long time since I did analysis. Could somebody show me how to show rigorously that $f(x)= |x|_\mathrm{eucl}^2$ is differentiable for all $x\in R^n$? I remember that the definition of differentiability involves if there exists a linear map $L$ s.t. ${|f(x+\epsilon)-f(x)-L(\epsilon)|\over|\epsilon|}\to0$ as $|\epsilon|\to0$ But is it necessary to find $L$ beforehand or is there some other way?

Sorry about this. Thank you in advance.

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You can invoke theorems: here, the partial derivatives of $f$ are all continuous, $\frac{\partial f}{\partial x_i} = 2x_i$, so the function is differentiable. –  Arturo Magidin Nov 21 '11 at 18:02
    
You do not have to use the definition. In your case it suffices to show that $f$ is continuously differentiable with respect to every coordinate. Try to write $f$ using such coordinates $x_1,\dots,x_n$. Calculating the partial derivatives should be easy after that. –  Matthias Klupsch Nov 21 '11 at 18:02
    
Thanks, Arturo and Matthias! –  hank Nov 21 '11 at 18:16

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up vote 7 down vote accepted

Going directly to the definition is unnecessarily complicated. The sane way to prove this is in steps:

  • Every constant function and each of the coordinate functions is differentiable (because it is a linear map).
  • Products and sums of differentiable functions are differentiable (with the usual product and sum rules; the definition can be used here, or you can reduce it to partial derivatives and use the equivalent result from single-variable analysis).
  • Every polynomial in the coordinate functions is differentiable (by the two previous points; no direct use of the definition).
  • Your $f(x)=|x|^2= x_1^2+x_2^2+\cdots+x_n^2$ is a polynomial in the coordinates and is therefore differentiable.
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THanks, this is MUCH better! –  hank Nov 21 '11 at 18:15

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