Can a function have partial derivatives, be continuous but not be differentiable?

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 differietable 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 $$(0,0)$$.

• Answer to title. Yes, it can. Sep 18 '20 at 10:44
• @AnginaSeng Thank you!
– 3ll
Sep 18 '20 at 10:52

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.

• Thank you for adding this!
– 3ll
Sep 19 '20 at 11:02

By differentiability theorem if partial derivatives exist and are continuos 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.

• (+1) All too easy Sep 18 '20 at 15:30
• Thanks Mark! With multivariable is less easy for me! Bye
– user
Sep 18 '20 at 15:35
• Thank you! And yes this is what I did to show it wasn't differentiable :)
– 3ll
Sep 19 '20 at 11:02
• You are welcome! Bye
– user
Sep 19 '20 at 11:16