Show that the function $f(x, y) = |xy|$ is differentiable at $(0, 0)$

Question

Firstly I understand that this question has already been asked : Show that the function $f(x,y) = |xy|$ is differentiable at 0, but is not of class $C^1$ in any neighborhood of 0. However that question did not address the question I'm about to ask.

---

In trying to show that $f : \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x, y) = |xy|$ is differentaible at $(0, 0)$ I first calculated the directional derivatives at $(0, 0)$ and I got $D_1f(0, 0) =D_2f(0, 0) = 0$, thus if $f$ is differentiable at $(0, 0)$ then we'd have $Df(0, 0) = [0 \ \ \ 0 ]$

Now I tried to prove that $f$ is differentiable at $(0, 0)$ by using the definition and $Df(0, 0)$ above as a canditate matrix for the derivative (because if the limit doesn't tend to zero using this matrix then it won't exist) and I arrived at the following (letting $h = (c, d)$)

$$\lim_{h \to (0, 0)} \frac{f((0, 0) + h) - f(0, 0) - Df(0, 0) \cdot h}{|h|} = \lim_{(c, d) \to (0, 0)} \frac{|cd|}{\sqrt{c^2 + d^2}}$$ but from the above it's not clear (to me) that $$\lim_{(c, d) \to (0, 0)} \frac{|cd|}{\sqrt{c^2 + d^2}} = 0$$

so I can't conclude that $f(x, y) = |xy|$ is differentiable at $(0, 0)$ using the definition. Now there's one other way I think I can prove differentiability which is by using the following theorem

Theorem: Let $A$ be open in $\mathbb{R}^m$. Suppose that the partial derivatives of the component functions of $f$ exist at each point $x$ of $A$ and are continuous on $A$. Then $f$ is differentiable at each point of $A$.

So I tried to show then that $D_1f(x, y)$ existed and was continuous on $\mathbb{R}^2$, in attempting to show this I ended up with the following

$$D_1f(x, y) = \lim_{t \to 0} \frac{f((x, y) + t(1, 0))-f(x, y)}{t} = \lim_{t \to 0}\frac{|(x+t)y|-|xy|}{t}$$ and again I'm not show how to evaluate this limit so I can't make use of the above theorem.

---

Is there any other way to show that $f$ is differentiable at $(0, 0)$? How could I show that $f$ is differentiable at $(0, 0)$?

Answer 1

$\displaystyle\frac {|cd|} {\sqrt {c^{2}+d^{2}}} \leq \frac {c^{2}+d^{2}} {2\sqrt {c^{2}+d^{2}}} \to 0$.

Answer 2

Note that $|x_k| \le \|x\|$, hence $|f(x)| \le \|x\|^2$.

Then $|f(x)-f(0) - 0. x| \le \|x\|^2$, hence $f$ is differentiable at $x=0$ with derivative $0$.

Answer 3

Note that, for $c\neq 0$, $d \neq 0$, (or, more compactly, $cd \neq 0$) $$ \sqrt{c^2 + d^2} = |c| \sqrt{\frac{c^2}{c^2} + \frac{d^2}{c^2}} = |cd| \sqrt{\frac{c^2}{c^2d^2} + \frac{d^2}{c^2d^2}} = |cd| \sqrt{\frac{1}{c^2} + \frac{1}{d^2}} \text{.} $$

Then $$ \frac{|cd|}{\sqrt{c^2 + d^2}} = \begin{cases} 0, & cd = 0 \\ \frac{1}{\sqrt{\frac{1}{c^2} + \frac{1}{d^2}}}, &cd \neq 0 \end{cases} \text{,} $$ both cases of which have limit $0$ as $(c,d) \rightarrow (0,0)$.

Answer 4

Note that

$$ \frac{|cd|}{\sqrt{c^2 + d^2}}\leq \frac{|cd|}{\sqrt{0+ d^2}}=|c| $$