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I have a question regarding the graphical intepretation of why the function $f(x, y) = |\ln(x) \cdot \ln(y)|$ is not differentiable on $(1, 1)$ :

3D plot of the function on Desmos 3D

I don't understand how from the plot we can see that on $(1, 1)$ the function is not differentiable ? Is there a trick like in single variable calculus when for example when we have an "edge" on a part of the plot of a function like in $(0, 0)$ for $|x|$ then we know the function is not differentiable at that point ?

Let me know if my question is not clear (I am new to this forum).

Thank you !

Sine of the Time
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rory
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  • How do you know it's not differentiable? – Sine of the Time Aug 10 '24 at 10:05
  • Do you know the result that relates differentiability in a point and the existence of the tangent plane? – Sine of the Time Aug 10 '24 at 10:14
  • If the graph of a function is not smooth or you can find any spikes or sharp edges it is not differentiable. But you can find a function which looks smooth but not differentiable. https://math.stackexchange.com/questions/3831023/can-a-function-have-partial-derivatives-be-continuous-but-not-be-differentiable – Madhan Kumar Aug 10 '24 at 10:14
  • @SineoftheTime According to wolfram alpha it isn't differentiable at point $(1, 1)$, but I don't know how to show it because starting from the definition, we have that f is differentiable at $(1, 1)$ if there exists a linear map $L_{(1, 1)} : \mathbb{R}^2 \mapsto \mathbb{R}$ and a fonction $r : E \mapsto \mathbb{R}$ such that $f(x, y) = f(1, 1) + L_{(1, 1)} ((x, y) - (1, 1)) + r(x, y)$ $\forall (x, y) \in E$ and $r$ must be such that $\lim_{(x, y) \rightarrow (1, 1)} \frac{r(x, y)}{||(x, y) - (1, 1)||} = 0$. So we need to compute this limit with $r(x, y) = f(x, y)$ but I don't know how – rory Aug 10 '24 at 11:39

2 Answers2

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First of all, if $f(x,y)$ is differentiable at $(1,1)$, it is obvious what the derivative must be.

When $x=1$, $f(x,y)=0$. And when $y=1$, $f(x,y)=0$. So any candidate for the linear function $L_{(1,1)}$ other than the zero function will result in a difference $r(x,y)$ such that $\frac{r(x,y)}{\|(x,y)-(1,1)\|}$ is a non-zero constant along each of these two lines (except at $(1,1)$, of course). So the derivative is either zero or nonexistent.

To prove that the derivative is zero, you could prove that both $\ln(x)\ln(y)$ and $-\ln(x)\ln(y)$ are differentiable at $(1,1)$, from which you could deduce the limit you need. Alternatively, you could use the squeeze theorem after finding suitable functions above and below $f(x,y)$.

A disproof could show two different limits when you approach $(1,1)$ along two different paths, or just show how to violate an arbitrary $\epsilon$ given an arbitrary $\delta$.

David K
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  • Thanks ! If I want to use the squeeze theorem would using $g(x, y) = (x-1)^2 + (y-1)^2$ work ? – rory Aug 11 '24 at 11:44
  • I think you would need to multiply that function by some factor greater than $1$, but that was the kind of function I was thinking of. – David K Aug 11 '24 at 15:29
  • Ok thanks ! But why do you said I would need to multiply by a factor greater than $1$ ? Because near $(1, 1)$ this function is an upper bound for $f$ isn't it ? – rory Aug 12 '24 at 10:06
  • Actually I’m not sure you need such a factor after all. It just seems easier to do the proof that way. You just have to be able to show that $f$ is less than the new function in some neighborhood of $(1,1)$. – David K Aug 12 '24 at 14:10
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We have $$ | \log a * \log | = |\log a | |\log b| $$

By $$\log x = \int_1^x \frac{1}{\xi} \ d\xi $$

we have an integral splitting for $x><1$ by the direction of

$$|\log x| = \cases{\int_1^x\ \frac{1}{\xi} \ d\xi \quad x>1\\ \int_x^1 \ \frac{1}{\xi} \ d\xi \quad x <1} $$

with a discontinuous derivative

$$\partial_x |\log x| = \cases{\frac{1}{x} \ \quad x>1\\ -\frac{1}{x} \quad x <1} $$

Since $x\to \log x, x>0$ is smooth, $(x,y) \to \log x \ \log y \quad x>0\ y>0$ is smooth in the first quadrant.

On the lines $x=1, y>0 $ and $y=1,x>0$ the function is zero with a change of sign.

The absolute value can be written by the product of signed value and the sign, that changes over the zero-lines

$$(x,y \to f(x,y) =|\log x \log y| = |\log x| | \log y| = \text{sign(x-1)} \text{sign(y-1)} \log x \log y $$

$$\frac{d}{dx} \left(\text{sign}(x-1) (x-1)\right) = 2 \ \delta(x-1)(x-1) + \text{sign}(x-1) 1 = \text{sign}(x-1)$$

So one gets the product of the sign functions at $(1,1)$ with a cross of non-differentiability along the two lines you see as a crossed kink line.

The point $1,1$ is the center of the cross with derivatives $\pm 1$ on its four quadrants.

Roland F
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  • Thanks for your answer ! But I don't understand the first part why you talk about $abs(x-1)$ and why you take the partial derivative of it. Could you elaborate more on that ? – rory Aug 10 '24 at 11:44
  • Certainly $|\ln(x)|$ has no derivative at $x=1$. But if you think about it very carefully, that doesn’t say much about the differentiability of $f$ at $(1,1)$. – David K Aug 11 '24 at 03:59
  • Thanks ! Could you just say why you take the partial derivative of $sign(x-1)(x-1)$ also I dont see why it is equal (for the first part of the sum) to $2 \delta(x-1)(x-1) + sign(x-1)1$

    Also what you get at the end is that on the cross the function is not differentiable so it is also not differentiable on $(1, 1)$ right ?

     – rory Aug 11 '24 at 11:36