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

• MHB
• mathmari
In summary: Yes, that is correct. You have shown that in the direction $(1,1)$, the limit does not exist, which means that the function is not differentiable in that direction.
mathmari
Gold Member
MHB
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:

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!

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.

Last edited:
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:

mathmari said:
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)

Klaas van Aarsen said:
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:

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

Klaas van Aarsen said:
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:

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)

Klaas van Aarsen said:
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:

mathmari said:
Ahh I see! But do we do that as I did it in my previous post with $(1,1)$ ?
For instance yes. (Nod)

Great! Thanks a lot! (Sun)

## What does it mean for a function to be partially differentiable in (0,0) but not totally differentiable?

Partial differentiability in (0,0) means that the function has a well-defined partial derivative at that point, in other words, the function is differentiable in that direction. However, not being totally differentiable means that the function is not differentiable in all directions at (0,0).

## Why is it important to distinguish between partial and total differentiability?

Distinguishing between partial and total differentiability is important because it allows us to understand the behavior of a function at a specific point. Partial differentiability gives us information about the slope of the function in a specific direction, while total differentiability gives us information about the overall behavior of the function.

## Can a function be partially differentiable but not continuous at (0,0)?

Yes, a function can be partially differentiable but not continuous at (0,0). This means that the function has a well-defined partial derivative at (0,0), but it has a discontinuity or a break in its graph at that point.

## What are some real-life examples of functions that are partially differentiable but not totally differentiable?

One example is a function that represents the temperature of a room. The temperature may be differentiable in the x-direction (left to right), but not in the y-direction (up and down). Another example is a function that represents the speed of a car. The speed may be differentiable in the direction of motion, but not in the opposite direction.

## How can we determine if a function is partially differentiable but not totally differentiable at a point?

To determine if a function is partially differentiable but not totally differentiable at a point, we can use the definition of partial derivatives. If the partial derivatives exist and are continuous at that point, then the function is partially differentiable. However, if the partial derivatives do not exist or are not continuous at that point, then the function is not totally differentiable.

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