Attempting to prove the differentiability of $f(x,y) = \vert xy \vert$

Question

We consider the function $f: \mathbb{R}^{2} \longrightarrow \mathbb{R}$, defined by $f(x, y):=|x y|$

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

I know this function appears a lot on this site but here is my attempt to proof a),b):

Approach to a):

To show the existence of $\frac{\partial f}{\partial x}(u), \frac{\partial t}{\partial y}(u)$, we need to show that $f$ is differentiable at $u \in U$.

Using $|x|=\sqrt{x^{2}},|y|=\sqrt{y^{2}}$ and the product rule, we obtain

$\frac{\partial f}{\partial x}(x, y)=f_{x}(x, y)=|y| \frac{x}{|x|}$ for $x=u_{1} \neq 0$

$\frac{\partial f}{\partial y}(x, y)=f_{y}(x, y)=|x| \frac{y}{|y|}$ for $y=u_{2} \neq 0$

Differentiability of $f$ at $(0, y)$ with $y \neq 0$ $\lim \limits_{h \rightarrow 0^{+}} \frac{f(0+h, y)-f(0,0)}{h}=\lim \limits_{h \rightarrow 0^{+}} \frac{|h||y|}{h}=|y|$ but $\lim \limits_{h \rightarrow 0^{-}} \frac{f(0+h, y)-f(0,0)}{h}=\lim \limits_{h \rightarrow 0^{-}} \frac{|h||y|}{h}=-|y|$

$\Rightarrow$ $f$ not differentiable at $(0, y)$ for $y \neq 0$ $\Rightarrow$ No partial derivatives of $f$ at $(0, y) \quad y \neq 0$

Differentiability of $f$ at $(x, 0)$ with $x \neq 0$ $\lim \limits_{h \rightarrow 0^{+}} \frac{f(x, h)-f(0,0)}{h}=\lim \limits_{h \rightarrow 0^{+}} \frac{|h||x|}{h}=|x|$ but $\lim \limits_{h \rightarrow 0^{-}} \frac{f(x, h)-f(0,0)}{h}=\lim \limits_{h \rightarrow 0^{-}} \frac{|h||x|}{h}=-|x|$

$\Rightarrow$ $f$ not differentiable at $(x, 0)$ for $x \neq 0$ $\Rightarrow$ No partial derivatives of $f$ at $(x, 0), x \neq 0$ Let $(x,y) = (0,0)$, then

$\begin{array}{l}\frac{\partial f}{\partial x}(0,0) =\lim \limits_{h \rightarrow 0} |0|\frac{h}{|h|}=0, \frac{\partial f}{\partial y}(0,0)=\lim \limits_{h \rightarrow 0}|0| \frac{h}{|h|}=0\end{array}$

So, $\frac{\partial f}{\partial y}(u), \frac{\partial f}{\partial x}(u)$ exist for $u=(0,0)$

It is probably still necessary to show that in all other points that do not lie on the axes, $f$ has partial derivatives.

Idea to fix this:

Can I simply argue that the remaining points are not "critical" points because, for example, the absolute value function $f(x) = |x|$ is differentiable for $x ≠ 0$? In this case, $f$ would not be differentiable if $x = 0$ or $y = 0$. But since I have already shown these cases, $f$ is differentiable for all $(x,y) \in \mathbb{R}^2 : (0,y), (x,0)$ with $x,y ≠ 0$ and therefore partial derivatives exist there.

Approach to b)

Differentiability of $f$ at $(0,0)$

$\begin{array}{l}\lim \limits_{\left(h_{1} ,h_{2}) \rightarrow(0,0)\right.} \frac{f\left(h_{1} h_{2}\right)-f(0,0)}{\|h\|} \=\lim \limits_{\left(h_{1},h_{2})\rightarrow(0,0\right)} \frac{\left|h_{1}\right|\left|h_{2}\right|}{\|h\|} \\ \leq \lim \limits_{\left(h_{1},h_{2}\right) \rightarrow(0,0)}\left|h_{1} h_{2} \right| =0\end{array}$

$\Rightarrow$ $f$ is differentiable at $(0,0)$

$f$ cannot be differentiable at $(0,y), (x,0)$ with $x,y ≠ 0$, since there are no partial derivatives there.

Therefore, $f$ is also differentiable at the points in a) where it has partial derivatives.

I have too many thoughts going through my head to put them all on paper. I know that the continuity and existence of partial derivatives at $u$ implies that $f$ is differentiable at $u$. But continuity is only a sufficient criterion. Maybe I still need it for b).

I am grateful for any help!

Answer 1

To show the existence of $\frac{\partial f}{\partial x}(u), \frac{\partial t}{\partial y}(u)$, we need to show that $f$ is differentiable at $u \in U$.

No, it is not needed for a function to be differentiable at a point $u$ for the partial derivatives to exist. There are non-differentiable functions with partial derivatives. However if $f$ is differentiable at $u$ then the partial derivatives must exist at $u$ (same goes for all directional derivatives at $u$) !

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

This is not using the definition of differentiability ! These are the partial derivatives so you should not say "differentiability of $f$ at ..." rather you should say "partial derivative of $f$ with respect to $x$ at ..." You are confusing the two meanings throughout the post.

an I simply argue that the remaining points are not "critical" points because, for example, the absolute value function f(x)=|x| is differentiable for x≠0 ? In this case, f would not be differentiable if x=0 or y=0 . But since I have already shown these cases, f is differentiable for all (x,y)∈R2:(0,y),(x,0) with x,y≠0 and therefore partial derivatives exist there.

It is probably still necessary to show that in all other points that do not lie on the axes, f has partial derivatives.

You've already proven that all partial derivatives exist at other points indeed you wrote

$\frac{\partial f}{\partial x}(x, y)=f_{x}(x, y)=|y| \frac{x}{|x|}$ for $x=u_{1} \neq 0$ $\frac{\partial f}{\partial y}(x, y)=f_{y}(x, y)=|x| \frac{y}{|y|}$ for $y=u_{2} \neq 0$

Then you write

Can I simply argue that the remaining points are not "critical" points because, for example, the absolute value function f(x)=|x| is differentiable for x≠0 ? In this case, f would not be differentiable if x=0 or y=0 . But since I have already shown these cases, f is differentiable for all (x,y)∈R2:(0,y),(x,0) with x,y≠0 and therefore partial derivatives exist there.

You have not proven that $f$ is differentiable ! You have only proven that the partial derivatives exist.

To prove that the function is differentiable use the fact that the composition of differentiable functions is differentiable. By using the function $x,y \mapsto xy$ and the absolute value function.