1. ## Differentiability

Hey everyone, I just want to make sure that I did this problem correctly. I have to show that this is not differentiable at the indicated point.

$f (x,y) = \sqrt{x^2+y^2} ; (0,0)$

And this is what I did:

$\partial f / \partial x = \frac{1}{2} (x^2 + y^2) ^\frac{-1}{2} = \frac{x}{\sqrt{x^2+y^2}}$

So at (0,0) this is undefined and it is the same for $\partial f / \partial y$. Thus is is not differentiable at the indicated point. Is this correct? Thanks for your help!

2. Hello,
Originally Posted by Layla
Hey everyone, I just want to make sure that I did this problem correctly. I have to show that this is not differentiable at the indicated point.

$f (x,y) = \sqrt{x^2+y^2} ; (0,0)$

And this is what I did:

$\partial f / \partial x = \frac{1}{2} (x^2 + y^2) ^\frac{-1}{2} = \frac{x}{\sqrt{x^2+y^2}}$

So at (0,0) this is undefined and it is the same for $\partial f / \partial y$. Thus is is not differentiable at the indicated point. Is this correct? Thanks for your help!
No.

For example, we have :

$\frac{\partial f}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$

If it is differentiable, then the limit as (x,y) approaches (0,0) must be the same, whatever direction you take.

In particular, $\lim_{x \to 0^+} \lim_{y \to 0} \frac{x}{\sqrt{x^2+y^2}}=\lim_{x \to 0^-} \lim_{y \to 0} \frac{x}{\sqrt{x^2+y^2}}$

But this is not true, as you can see if you calculate the limit $\frac{x}{\sqrt{x^2}}$

3. Originally Posted by Layla
So at (0,0) this is undefined and it is the same for $\partial f / \partial y$. Thus is is not differentiable at the indicated point. Is this correct?
Not quite. What you proved is that $\frac{\partial f}{\partial x}$ cannot be continuously extended at 0 (the limit depends on the sign of $x$ when it approaches 0).
The following works: if $f$ were differentiable at $(0,0)$, then the function $x\mapsto f(x,0)=|x|$ would be differentiable at $0$ (it is just saying that if $f$ is differentiable at $(0,0)$, then $\frac{\partial f}{\partial x}$ exists at (0,0), since this is a "directional derivative" at $(0,0)$). However, as you know, this is false.

4. Originally Posted by Moo
$\frac{\partial f}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$

If it is differentiable, then the limit as (x,y) approaches (0,0) must be the same, whatever direction you take.
The function $f:x\mapsto x^2\sin\frac{1}{x}$ is continuous at 0, $f'$ is not continuous at 0 (as you can check by computation). However, $f$ is differentiable at 0 ( $\frac{f(x)}{x}\to 0$ as $x$ tends to 0)...

Your argument could be made rigorous using additional properties. You can find two directions ( $x\mapsto(\pm x,0)$ for $x>0$) where different finite limits exist for $\frac{\partial f}{\partial x}$. And then use the following fact: if a continuous function is differentiable on $(a,b)$ and such that its derivative admits a finite limit $\ell$ at $b$, then the function is differentiable at $b$ and its derivative at $b$ is $\ell$. But this is a bit overcomplicated.

5. Originally Posted by Laurent
The function $f:x\mapsto x^2\sin\frac{1}{x}$ is continuous at 0, $f'$ is not continuous at 0 (as you can check by computation). However, $f$ is differentiable at 0 ( $\frac{f(x)}{x}\to 0$ as $x$ tends to 0)...

Your argument could be made rigorous using additional properties. You can find two directions ( $x\mapsto(\pm x,0)$ for $x>0$) where different finite limits exist for $\frac{\partial f}{\partial x}$. And then use the following fact: if a continuous function is differentiable on $(a,b)$ and such that its derivative admits a finite limit $\ell$ at $b$, then the function is differentiable at $b$ and its derivative at $b$ is $\ell$. But this is a bit overcomplicated.
Awww... My remnants of multivariable calculus look far behind lol! Thanks for your input

(I think I was trying to do the thing you described in the latter part ^^)