Differentiability of $f(x,y) = |xy|$

Question

I know there is also a questions concerning this task. However I have some questions about my approaches to solve this taks. Thanks for your help in advance!

We consider the function $f: \mathbb{R}^{2} \longrightarrow \mathbb{R}$, defined by $$f(x, y):=|x y|$$ a) Determine all points $u=(u_{1}, u_{2}) \in \mathbb{R}^{2}$ at which the function $f$ has partial derivatives $\frac{\partial f}{\partial x}(u)$ and $\frac{\partial f}{\partial y}(u)$.

b) Determine all points $u=(u_{1}, u_{2}) \in \mathbb{R}^{2}$ at which the function $f$ is differentiable.

Question to a)\ When I compute the partial derivatives for $\frac{\partial f}{\partial x}(u)$ and $\frac{\partial f}{\partial y}(u)$ I have

$$\frac{\partial f}{\partial x}(u) = \frac{|y|x}{|x|}$$ and $$\frac{\partial f}{\partial y}(u) = \frac{|x|y}{|y|}$$

We see that we can't have $(0,y) \in \mathbb{R}^{2}$ for $\frac{\partial f}{\partial x}(u)$ and $(x,0)\in \mathbb{R}^{2}$ for $\frac{\partial f}{\partial y}(u)$. And the partial derivatives also don't exist in $(0,0) \in \mathbb{R}^{2}$.

I feel like this doesn't solve the task correctly. Is there anything I'm missing?

Question to b)\ b) In order to find differentiability for $f(x,y)$ in $u=(u_1,u_2)$ we have to find the points where the partial derivatives exist and are continuous. So we have to deal with our special points $(0,0),(x,0),(0,y) \in \mathbb{R}^{2}$

1. Let $(x,y)=(0,0)$, then

$$\begin{aligned}\lim \limits_{(h_{1},h_{2}) \rightarrow(0,0)} \frac{f(h_{1}, h_{2})-f(0,0)}{\|h\|}&=\lim \limits_{(h_{1},h_{2})\rightarrow(0,0)} \frac{|h_{1}| \cdot|h_{2}|}{\sqrt{h_{1}^{2}+h_{2}^{2}}} \\ &\leq \lim \limits_{(h_{1},h_{2}) \rightarrow(0,0)}|h_{1}| \cdot|h_{2}| =0\end{aligned}$$

So $f(x,y)$ is differentiable in $u=(0,0)$.

2. Let $(x,y)= (0,y)$, then

$$\lim \limits_{h \rightarrow 0^{+}} \frac{f(h, y)-f(0, y)}{h}=\lim \limits_{h \rightarrow 0^{+}} \frac{|h||y|}{h}=|y|$$

but

$$\lim \limits_{h \rightarrow 0^{-}} \frac{f(h, y)-f(0, y)}{h}=\lim \limits_{h \rightarrow 0^{-}} \frac{|h||y|}{h}=-|y|$$

So $f$ is not differentiable in $(0,y) \in \mathbb{R}^{2}$.

3. Let $(x, y)=(x, 0)$, then

$$\lim \limits_{h \rightarrow 0^{+}} \frac{f(x, h)-f(x, 0)}{h}=\lim \limits_{h \rightarrow 0^{+}} \frac{|h||x|}{h}=|x|$$

but

$$\lim \limits_{h \rightarrow 0^{-}} \frac{f(x, h)-f(x, 0)}{h}=\lim \limits_{h \rightarrow 0^{-}} \frac{|h||x|}{h}=-|x|$$

So $f$ is not differentiable in $(x, 0) \in \mathbb{R}^{2}$.

So $f$ is $\forall(x, y) \in \mathbb{R}^{2} \backslash((x, 0),(0, y))$ differentiable.

Let me know what you think, thanks for your help!

Answer 1

a) You did not completely solve (a) but you did most of the work. You:

• calculated $\frac{∂f}{∂x}(u)$ (resp. $\frac{∂f}{∂y}(u)$) when $u_1\ne0$ (resp. $u_2\ne0$);
• (in b2 and b3 but that should take place in a) proved that (if $u\ne(0,0),$ which you forgot to mention) $\frac{∂f}{∂x}(u)$ (resp. $\frac{∂f}{∂y}(u)$) does not exist when $u_1=0$ (resp. $u_2=0$) (btw, here is a duplicate);
• claimed (wrongly, since you stated in b1 that $f$ is differentiable at $(0,0)$) that $\frac{∂f}{∂x}(0,0)$ and $\frac{∂f}{∂y}(0,0)$ don't exist.

b)

• At $(0,0),$ your proof in b1 that $f$ is differentiable is (also a duplicate and) fine but you forgot to say what the differential is.
• At $(u_1,0)$ or $(0,u_2)$ except at $(0,0),$ as previously said, your b2 and b3 should take place in a, and here you can simply derive that since the partial differential do not both exist, $f$ is not differentiable.
• On the four open quadrants, you forgot to state that $f$ is differentiable.
  • It is even continuously differentiable, since the partial derivatives calculated in (a) are continuous on these open subsets. Beware that the continuity of the partial derivative is a sufficient condition of differentiability, but not a necessary one as you claimed: " In order to find differentiability for $f(x,y)$ in $u=(u_1,u_2)$ we have to find the points where the partial derivatives exist and are continuous. ".
  • Alternatively , these four restrictions are even polynomial (either $=xy$ or $-xy,$ depending on which quadrant) hence smooth.