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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 $0$ ?

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  • $\begingroup$ 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 $(0,0)$. $\endgroup$ – Indrajit Ghosh Sep 1 '18 at 7:44
  • $\begingroup$ @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? $\endgroup$ – jeea Sep 1 '18 at 10:03
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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.

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

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