10
$\begingroup$

I have a function:

$$ f(x,y)= \begin{cases} \dfrac{2x^2y+y^3}{x^2+y^2} & \text{if $(x,y) \neq (0,0)$}\\ 0 & \text{if $(x,y) = (0,0)$}\\ \end{cases} $$

which I think I managed to show:

a) continuity at $(0,0)$
by $\lim_{(x,y) \to (0,0)} f(x,y) = 0$

b) has partial derivatives at $(0,0)$
by the definition of derivatives and found $f'_x(0,0) = 0, f'_y(0,0) =1$. Still not 100% sure if did this correctly.

c) not differentiable at $(0,0)$
by definition of differentiable functions and that a limit didn't exist.

However, I feel like because of this I can tell more about the function. I'd like it if someone can confirm this. I assumed, that because it wasn't differentiable, the partial derivatives might not be continuous around $(0,0)$. $$\frac{\partial f}{\partial x} = \frac{2y^3x}{\left(x^2+y^2\right)^2}$$ $$\frac{\partial f}{\partial y} = \frac{y^4+y^2x^2+2x^4}{\left(x^2+y^2\right)^2}$$

Is that the case? I checked the limits $$\lim_{(x,y) \to (0,0)} \frac{\partial f}{\partial x} \quad \text{and} \quad \lim_{(x,y) \to (0,0)} \frac{\partial f}{\partial y}$$ and they don't seem to exist. What would happen if one existed but not the other? Is this possible? What would happen if the limit was something else than $0$ and $1$ I calculated in b)? Just not being continuous? I am just worried if the function really has partial derivatives in $(0,0)$.

Thank you in advance!

$\endgroup$
3
  • $\begingroup$ Answer to title. Yes, it can. $\endgroup$ Sep 18, 2020 at 10:44
  • $\begingroup$ @AnginaSeng Thank you! $\endgroup$
    – lotus57
    Sep 18, 2020 at 10:52
  • $\begingroup$ Is the function weakly differentiable? Is $f\in W^{1,1}([0,1]^2)$? 🤔 $\endgroup$ Feb 9, 2023 at 13:27

2 Answers 2

16
$\begingroup$

To complement user's answer, I would like to point out that the example in the OP is even more striking since not only do partial derivatives $\partial_1f(0,0)$ and $\partial_2f(0,0)$ exists, but also the directional derivative of the function $f$ at $\boldsymbol{0}=(0,0)$ along any direction $\mathbf{v}=(h,k)$ exists:

$$\partial_\mathbf{v}f(0,0):=\lim_{t\rightarrow0}\frac{f(\boldsymbol{0}+t\mathbf{v})-f(\boldsymbol{0})}{t}=\lim_{t\rightarrow0}\frac{1}{t}\frac{t^3k(h^2+k^2)}{t^2(h^2+k^2)}=k$$

So to add to other solutions:

A function $f$ may be

  • continuous at some point $\mathbf{c}$,
  • have (finite) directional derivatives at $\mathbf{c}$ along any vector $\mathbf{v}$ ($\partial_1f(\mathbf{c})$ and $\partial_2f(\mathbf{c})$ correspond to $\mathbf{v}=(1,0)$ and $\mathbf{v}=(0,1)$ respectively)

and yet not be differentiable.

$\endgroup$
1
  • $\begingroup$ Thank you for adding this! $\endgroup$
    – lotus57
    Sep 19, 2020 at 11:02
10
$\begingroup$

By differentiability theorem if partial derivatives exist and are continuous in a neighborhood of the point then (i.e. sufficient condition) the function is differentiable at that point.

The existence of partial derivatives doesn't suffice.

In this case we can proceed by definition

$$ \lim_{(h,k) \to (0,0)} \frac{\dfrac{2h^2k+k^3}{h^2+k^2}-k}{\sqrt{h^2+k^2}}= \lim_{(h,k) \to (0,0)} \frac{2h^2k+k^3-kh^2-k^3}{\sqrt{(h^2+k^2)^3}}= \lim_{(h,k) \to (0,0)} \frac{h^2k}{\sqrt{(h^2+k^2)^3}}$$

which doesn't exist and therefore the function is not differentiable at $(0,0)$ and indeed partial derivatives are not continuous at that point.

$\endgroup$
5
  • 1
    $\begingroup$ (+1) All too easy $\endgroup$
    – Mark Viola
    Sep 18, 2020 at 15:30
  • $\begingroup$ Thanks Mark! With multivariable is less easy for me! Bye $\endgroup$
    – user
    Sep 18, 2020 at 15:35
  • $\begingroup$ Thank you! And yes this is what I did to show it wasn't differentiable :) $\endgroup$
    – lotus57
    Sep 19, 2020 at 11:02
  • $\begingroup$ You are welcome! Bye $\endgroup$
    – user
    Sep 19, 2020 at 11:16
  • 1
    $\begingroup$ Actually, for differentiability only at a certain point, continuity of the partial derivatives at that point is sufficient. If you have continuity of the partial derivatives at a whole open set containing the point, then the function is already $C^1$ in that open set, which is stronger than just being differentiable at a point. $\endgroup$ Dec 31, 2023 at 11:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .