# At which points is this function differentiable

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 $x_i=0$?