# Non differentiability of $|x| + |y|$

I have to show non differentiability of $f(x,y) = |x| + |y|$ at $(0,0)$.

Now we know this is continuous at $(0,0)$, so I tried finding $f_x$ and $f_y$ using the theorem, which says that if $f(x,y)$ is differentiable at $(0,0)$, then

$\Delta f = f(x+h, y+k)- f(x,y) = hf_x(0,0) + kf_y(0,0) + \epsilon_1 h + \epsilon_2 k$

where $\epsilon_i$ are function of $h,k$ and Both $\to 0$ as $h,k \to 0$. Now finding it here, we have:

$$\Delta f = |h| + |k|\tag{1}$$

comparing, i took $\epsilon_i = 0$ and $f_x(0,0)$ and $f_y(0,0)$ depend on path whether $h,k > 0$ or $< 0$. That is $f_x(0,0)$ is $1$ for $h>0$ and $-1$ for $h< 0$.

1. Can we now say that function is non-differentiable as $f_x$ and $f_y$ dont exist at $(0,0)? 2. Actually by checking$h$and different value of$f_x$have I checked existence of$f_x(0,0)$or continuity of$f_{x}(x,y)$at$(0,0)$? I think I have done the former. 3. How do we know that in$(1)$we have to take$\epsilon_1$as coefficient of$|h|$or as$0$? • If a function$f:\Bbb{R}^2 \to \Bbb{R}$is differentiable at$a$then$f_x(a)$and$f_y(a)$must exist....So as u see that this function does not have the partial derivatives at$(0,0)$...so it is not differentiable at$(0,0)$. – Indrajit Ghosh Sep 1 '18 at 7:44 • @IndrajitGhosh Yes i have many questions in this regard, for example, the question 2, what am I really checking? Is it continuity that I checked, or is it existence of partial at origin? – jeea Sep 1 '18 at 10:03 ## 1 Answer You should check that the partial derivatives at the origin exist before looking at$\Delta f$. In this case, since$f(x,0)=|x|$is not differentiable at$x=0$(in the one-variable sense), the partial derivative$f'_x(0,0)$doesn't exist. Therefore$f\$ isn't differentiable (in the two-variable sense) at the origin.

• Thanks this makes it simpler to use! – jeea Sep 2 '18 at 8:45