F is partially differentiable in (0,0) but not total differentiable

[mathmari]

Hey! :giggle:

Let's consider the function \begin{align*}f:\mathbb{R}^2&\rightarrow \mathbb{R} \\ (x,y)&\mapsto \sqrt{|x|\cdot |y|}\end{align*}
Show that $f$ is partially differentiable in $(0,0)$ but not total differentiable.I have done the following:
We prove that $f$ is partially differentiable in each direction:
Let $0 \neq v =(r, s) \in \mathbb{R}^2$ be the direction and $0 \neq h \in \mathbb{R}$ :
\begin{align*}\frac{1}{h}\cdot \left (f\left ((0,0)+hv\right )-f(0,0)\right )&=\frac{1}{h}\cdot \left (f(hr,hs)-0\right )=\frac{1}{h}\cdot f(hr,hs)=\frac{1}{h}\cdot \sqrt{|hr|\cdot |hs|}=\frac{1}{h}\cdot \sqrt{|h|^2\cdot |r|\cdot |s|}\\ &=\frac{1}{h}\cdot |h|\cdot \sqrt{|r|\cdot |s|} =\begin{cases}\frac{1}{h}\cdot h\cdot \sqrt{|r|\cdot |s|} & \text{ if } h\geq 0\\ \frac{1}{h}\cdot \left (-h\right )\cdot \sqrt{|r|\cdot |s|} & \text{ if } h< 0\end{cases}\\ &=\begin{cases}\sqrt{|r|\cdot |s|} & \text{ if } h\geq 0\\ - \sqrt{|r|\cdot |s|} & \text{ if } h< 0\end{cases}\end{align*}
Since this term is independent from $h$, the limit for $h \rightarrow 0$ is equal to $\pm \sqrt{|r|\cdot |s|}$,which means that the limit exists.
That means that $f$ is partially differentiable in every direction, since for each $v$ there is a directional derivative.

Is that correct? Or do we show that $f$ partially differentiable by calculating the partial derivatives $f_x$ and $f_y$ ? :unsure:

To show that that $f$ is not totally differentiable do we show that that $f$ is not continuous in $(0,0)$ ? :unsure:

 

[HOI]

So a function of two or more variables is "partially differentiable" at a point if its partial derivatives exist there?
(I don't believe that is stanard terminology. I would just say "has partial derivatives".)

There is a theorem that says that a function, f, of two or more variables is "differentiable" at a point if and only if its partial derivative exist and are continuous there (just "f is continuous" is not sufficient).

Here, $f(x,y)= \sqrt{|x||y|}= (|x||y|)^{1/2}$.

The partial deriatives, at (0,0), are particularly easy. To find the partial derivative with respect to x, at (0,0) we take the derivative with respect to x while holding y= 0. Then $f(x, 0)= (|x|(0))^{1/2}= 0$ for all x. That is constant so the partial derivative with respect to x is 0. The same is true of the partial derivative with respec to y.

To find the total derivative at (0,0) we would have to have a neighborhood of (0,0) in which we could calculate $\frac{f(x, y)- f(0, 0)}{\sqrt{x^2+ y^2}}$. But any such neighborood woud include points in the second and fourth quadrants where xy is negative and the function value does not even exist!

 

[I like Serena]

Partial differentiable just means that the partial derivatives exist. That is different from differentiable in a specific direction, which is again different from (totally) differentiable.

Using your argument, we can substitute r=1 and s=0 to see that $f_x$ exists and is 0 at (0,0).

Using your argument, we can also pick a direction where both r and s are non-zero and see that we get a different limit depending on the sign of h. Therefore there is a direction in which f is not differentiable, which implies that f is not totally differentiable either.

Btw, isn't f continuous?
Then we can't use that as an argument.

 

[mathmari]

So to show that $f$ is partially differentiable we just have to calculate the partial derivatives?

For that do we do the following?

Let $h \in \mathbb{R}\setminus \{0\}$ and $j \in \{1, 2\}$. Then we get \begin{equation*}\frac{f\left ((0,0)+he_j\right )-f(0,0)}{h}=\frac{f\left (he_j\right )-0}{h}=\frac{f\left (he_j\right )}{h}\end{equation*}
For $j=1$ :
\begin{equation*}\frac{f\left (he_1\right )}{h}=\frac{f\left (h(1,0)\right )}{h}=\frac{f(h,0)}{h}=0\rightarrow 0\end{equation*}
For $j=2$ :
\begin{equation*}\frac{f\left (he_2\right )}{h}=\frac{f\left (h(0,1)\right )}{h}=\frac{f(0,h)}{h}=0\rightarrow 0\end{equation*}
Therefore $f$ is in $(0, 0)$ partially differentiable with \begin{equation*}\frac{\partial{f}}{\partial{x}}(0,0)=\frac{\partial{f}}{\partial{y}}(0,0)=0\end{equation*}

Is that correct? Or have I understand that wrongly?

:unsure:

 

[I like Serena]

Therefore $f$ is in $(0, 0)$ partially differentiable with \begin{equation*}\frac{\partial{f}}{\partial{x}}(0,0)=\frac{\partial{f}}{\partial{y}}(0,0)=0\end{equation*}

Correct. (Nod)

 

[mathmari]

Correct. (Nod)

So to show that $f$ partially differentiable in $(0,0)$ do we need to do only what I have done in post #4 ? Or do we have to do something further? Do we need the part I wrote in post #1 ? :unsure:

 

[I like Serena]

Yes. Post 4 covers partial differentiability.
That leaves total differentiability for which we can use post 1 to show f is not differentiable..

 

[mathmari]

That leaves total differentiability for which we can use post 1 to show f is not differentiable..

For that we have to pick a specific direction $(r,s)$ with $r,s\neq 0$, or not?
For example $r=s=1$ then we have that
\begin{align*}\frac{1}{h}\cdot \left (f\left ((0,0)+h(1,1)\right )-f(0,0)\right )&=\frac{1}{h}\cdot \left (f(h,h)-0\right )=\frac{1}{h}\cdot f(h,h)=\frac{1}{h}\cdot \sqrt{|h|^2}\\ &=\frac{1}{h}\cdot |h| =\begin{cases}\frac{1}{h}\cdot h & \text{ if } h\geq 0\\ \frac{1}{h}\cdot \left (-h\right )& \text{ if } h< 0\end{cases}\\ &=\begin{cases}1 & \text{ if } h\geq 0\\ - 1 & \text{ if } h< 0\end{cases}\end{align*}
Since for $h\rightarrow 0^+$ we get $1$ and for $h\rightarrow 0^-$ we get $-1$, this means that we don't have continuity there and so $f$ cannot be differentiable.

Is that correct? :unsure:

 

[I like Serena]

It's not about continuity.
Instead we have that the limit does not exist to find a directional derivative.
And if a directional derivative does not exist, then that implies that the function is not differentiable. (Nerd)

 

[mathmari]

It's not about continuity.
Instead we have that the limit does not exist to find a directional derivative.
And if a directional derivative does not exist, then that implies that the function is not differentiable. (Nerd)

Ahh I see! But do we do that as I did it in my previous post with $(1,1)$ ? :unsure:

 

[I like Serena]

Ahh I see! But do we do that as I did it in my previous post with $(1,1)$ ?

For instance yes. (Nod)

 

[mathmari]

Great! Thanks a lot! (Sun)