I am currently learning about differentiability of multi variable functions. In my textbook, they use the example of the function $f(x,y)=||x|-|y||-|x|-|y|$. The partial derivatives are demonstrated to be $f_x=0$ and $f_y=0$. The book continues on to use the equation of the tangent plane $z=f(a,b)+f_x(a,b)(x-a) + f_y(a,b)(y-b)$ to state that the tangent plane for this function must be $z=0$.

The book proves through one definition (checking of the tangent place is a 'good' approximation using limits) that the function is not differentiable at $(0,0)$ and this tangent plane is not valid which makes sense to me. However, where I get stumped is the next definition of differentiability the book provides.

It provides a theorem which states: "Suppose $X$ is open in $R^2$. If $f:X→R$ has continuous partial derivatives in a neighborhood of $(a,b)$ in $X$, then $f$ is differentiable at $(a,b)$."

By this definition should this function not be differentiable?

• Its domain is open in $R^2$
• Its partial derivatives $f_x,f_y=0$ are continuous everywhere including $(0,0)$

But it was already shown that the function is actually not differentiable at $(0, 0)$. What am I missing? I feel like something simple has slipped by me.

• m.wolframalpha.com/input/…. Partial derivatives are $0$ ONLY aren't zero always. In particular they arent contuous, so there is no contradiction here. – Dog_69 Jun 12 '18 at 12:38

Why do you say that $f_x=f_y=0$? This is true at $(0,0)$, indeed, but not everywhere.
For instance, near $(2,1)$, we have$$f(x,y)=\bigl||x|-|y|\bigr|-|x|-|y|=x-y-x-y=-2y.$$Therefore, $f_y(2,1)=-2$.
• I think this might be it. Looking again at my textbook it looks like I accidentally thought $f_x(0,0)$ was actually $f_x$ in general. Thanks for clearing it up. – Dan Jun 12 '18 at 12:39
The partial derivates don't exist on an open neighborhood of $(0,0)$. You can check that they don't exist on $$\{(x,x)\in\mathbb R^2~:~x\neq 0\}.$$