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Find all points $(x_1,x_2,\ldots,x_k) = x \in \mathbb{R}^k$ such that $f(x)=|x_1 \cdot x_2 \cdots x_k|$ is a differentiable function.

I know that if $\forall i, x_i \neq 0$ then this function is differentiable. I don't know what about rest of points.

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Let $(x_1,\ldots,x_k)$ be a point where $f(x)=0$. If $x_i=0$ then $\frac\partial{\partial x_j}f(x)=0$ for all $j\ne i$. Thus if exactly one coordinate are zero all other partial derivatives are zero, while for the $i$th component we have that $f$ looks like $c|x_i|$, where $c$ is the (nonzero) absolute value of the rproduct of all other coordinates, hence $\frac\partial{\partial x_i}f(x)$ does not exist and $f$ is not differentiable.

But if $r\ge 2$ coordinates are zero, all partial derivatives are zero. In fact, we can see that $\frac{f(x+h)}{||h||}\to 0$ as $h\to 0$:

  • If $r=k$, then $|f(x+h)|\le |h|_\infty^k\le ||h||^k$ and from $k= r\ge 2$, the claim follows.
  • If $r<k$, let $m=\min\{|x_i|\colon x_i\ne 0\}$. Then for $||h||<m$, we have $f(x+h)\le |h|_\infty^r (2m)^{k-r}$, whence again the claim.

Thus $f$ is totally differentiable (with derivative 0) if at least two coordinates are zero.

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$f$ is differentiable at $x$ if and only if all its partial derivatives exist at $x$. If you vary $x_i$ while keeping the other components fixed, is the resulting function $f_i(x_i)$ of one variable differentiable at $x_i=0$?

If you do this, you will see that the function is differentiable only if all its variables are nonzero. Special care need to be given to the points where more than one coordinate is zero, but it is not differentiable there either, by a well-definedness-argument.

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Your second paragraph contradicts the first: At points with multiple zeroes, all partials are zero. – Hagen von Eitzen Oct 25 '12 at 18:01

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