Derivates of $f(x,y)=|x| + |y| − ||x| − |y||.$

Question

I worked on one my problem and I want you to check my work or give my some good idea for solution. $$f(x,y)=|x| + |y| − ||x| − |y||.$$ a) Find all $(x,y)$ such that $f$ is continuous at given $(x,y)$.

Solution a) it's very easy to see that given function is continuous at every $(x,y)$ because absolute value function $|x|$ is continuous at every $x$.

b) Count partial derivatives of function $f$ at the point $(0,0).$

Solution: My idea is not to get partial derivatives for every $x$ and $y$, but just use limit definition of it.

$$\lim_{h\to0} [\frac{f(h,0)-f(0,0)}{h}=\frac{|h|+|0|-||h|-|0||}{h}=\frac{0}{h}=0]$$ This is partial derivatives whith respect of x. Similarly we get $0$ for partial derivatives with respect to y at the point $(0,0)$.
c) Is it function $f$ differentiable at $(0,0)$?

My work: Idea is to use definition, so we know how looks our differential operator - it's null operator (from b) part).
$$\lim_{(x,y)\to0} \frac{f(x,y)-f(0,0)-A(x,y)}{||(x,y)||}=\frac{|x| + |y| − ||x| − |y||}{\sqrt{(x^2+y^2)}}$$. If we look at $x=0$ limit is $0$, but if $y=x$ than limit isn't equal to zero, so this limit doesn't exist.

Answer 1

As you may be aware, if $a$ and $b$ are real numbers, then \begin{align*} \max(a, b) &= \tfrac{1}{2}(a + b + |a - b|), \\ \min(a, b) &= \tfrac{1}{2}(a + b - |a - b|). \end{align*} (If $I$ denotes the closed interval with endpoint $a$ and $b$, each formula may be interpreted as "starting at the midpoint of $I$ and traveling to the right (max) or left (min) by half the length of $I$.)

Particularly, $$ f(x, y) = |x| + |y| - \bigl||x| - |y|\bigr| = 2\min(|x|, |y|). $$ This observation makes your conclusions geometrically apparent.

Answer 2

Your work looks very good. Assuming this is homework, the only thing I would add is a bit more information about the very last statement. That the limit is $0$ if $x = 0$ is evident but you might show a little more explicitly why the limit is not $0$ if $y = x$.

Answer 3

Just a tip / hint.
An isoline chart / contour plot of the function at hand looks like this; darker means lower values.