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I need an example of a function $\displaystyle f:\mathbb R^2 \rightarrow \mathbb R^2$ such that $f$ is differentiable everywhere except at the point $(1, 2)$.

I was thinking (similar to single variable functions) about using the absolute value, for an example $\displaystyle f(x,y) = (|x-1|, |x-2|)$ but I'm not sure if that is correct. Could, as an analogy, a norm be used? Something like $f(x,y)= ||(x-1,y-2)||$

Thanks for your help!!

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    I think I would take $z=\sqrt{x^2+y^2}$ and move the vertex where I wanted it. – 2017-01-21

2 Answers 2

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A norm cannot be used, since that is a function $\mathbb R^2 \rightarrow \mathbb R$, not $\mathbb R^2 \rightarrow \mathbb R^2$.

Your example $f(x,y)=(|x−1|,|x−2|)$ is not differentiable in $\{x=1\}\cup\{y=2\}$.

The easiest way is to make an artificial discontinuity:

$$f(x,y) := \left\{\begin{matrix} (0,0) \text{ if } (x,y) \ne (1,2)\\ (1,1) \text{ if } (x,y) = (1,2) \end{matrix}\right.$$

If $f$ is not continuous at a certain point, it can't be differentiable there.

2

Absolutely yes.

The function you gave is great since the euclidean norm on $\mathbb R^2$ is differentiable on $\mathbb R^2\setminus\{0\}$.

Your example

$$f(x,y)=\vert\vert(x-1,y-2)\vert \vert$$

works.