# What is the meaning of “has partial derivatives everywhere”?

I am reading Courant's "Differential and Integral Calculus Vol.2".

I am confused with something: Courant says that the function defined as $u(x,y)=\frac{2xy}{x^2 +y^2}$ and $u (0,0)=0$ is not continuous but has partial derivatives everywhere.

I am assuming that "has partial derivatives everywhere" means "is differentiable". I could be wrong.

I've seen somewhere else that a function is differentiable at $(x_0, y_0)$ if there is $(\alpha_1, \alpha_2)\in \mathbb{R}^2$ such that:

$$\lim_{(h,k)\to (0,0)} \frac{f(x_0+ h, y_0 + k )-f(x_0,y_0) - \alpha_1 h - \alpha_2 k }{\sqrt{h^2 + k^2}}=0$$

Applying to our function, we have:

$$\lim_{(h,k)\to (0,0)} \frac{2hk}{(h^2 + k^2)\sqrt{h^2 + k^2}}-\frac{\alpha_1 h +\alpha_2 k}{\sqrt{h^2 + k^2}}$$

With polar coordinates $h\to r\cos, k\to r \sin$.

$$\lim_{r\to 0} \frac{2\cos \theta \sin \theta}{r}- \alpha_1 \cos \theta - \alpha_2 \sin \theta = \lim_{r\to 0} \frac{\sin{2\theta}}{r} - \alpha_1 \cos \theta - \alpha_2 \sin \theta$$

I guess this shows that the limit depends on $\sin 2\theta$ being equal to $0$ and hence can not exist. So what Courant meant with "partial derivatives existing everywhere"? It could be $\mathbb{R}\setminus\{(0,0)\}$ but I wouldn't call this "everywhere".

Also, when defining $u(0,0)=0$, what does this means for the partial derivatives? What does this means for $u_x(0,0)$ and $u_y(0,0)$?

• Well, it could be differentiable, but the claim that it has partial derivatives everywhere is a different claim. Partial derivatives leave one of the variables constant, equal to the corresponding coordinate of the points at which the partial derivative is being taken. Differentiable is the defined as holding the first displayed equation that you wrote for some constants $\alpha_1,\alpha_2$. – user568248 Jun 10 '18 at 0:37
• Partial derivatives need to satisfy additional constraints in order for the function to be differentiable, such as the Cauchy-Riemann Equations. – CyclotomicField Jun 10 '18 at 0:46
• @CyclotomicField Careful! Cauchy-Riemann equations holding at a point don't imply differentiability. Strictly speaking, they are also out of the scope here. They make sense for functions $\mathbb{R}^2\to\mathbb{R}^2$. You are confusing differentiability with complex-differentiable. And even in that case, the components of the function need to be differentiable for the theorem to hold. – user568248 Jun 10 '18 at 0:56
• @wakandivostok I'm not confusing I just felt it was an accessible example of when additional constraints are required to ensure differentiability. It's certainly a little out of scope but it's an easy thread to tug on so I thought it was worth mentioning anyway. – CyclotomicField Jun 10 '18 at 1:39
• @CyclotomicField The thing is that Cauchy-Riemann equations is not a condition that, if added, implies differentiability. So, it is questionable how that is an example of an additional constrain required. It is like if one said "... satisfy additional constrains, ... such as being periodic". It is an extra condition, but it doesn't make the function differentiable. – user568248 Jun 10 '18 at 1:43

The function $u$ clearly has partial derivatives at all points $\neq (0,0)$, since it is an elementary function in that open domain. At $(0,0)$ the partial derivatives would be

$$u_x(0,0)=\lim_{x\to0}\frac{u(x,0)-u(0,0)}{x}=\lim_{x\to0}\frac{0}{x}=0$$

Likewise

$$u_y(0,0)=\lim_{y\to0}\frac{u(0,y)-u(0,0)}{y}=\lim_{y\to0}\frac{0}{y}=0$$

Therefore, the partial derivatives exist everywhere.

It looks then, that the point of the exercise is to show that differentiability and existence of partial derivatives are not equivalent. This rounds up the relation between these two concepts, since differentiability at a point, implies that the partial derivatives exist.

• Oh, then they are two different things! Thanks, man. I was going insane here. I also checked the partial derivatives and obtained that result, but when I went to check for differentiability, things got weird because I thought they were the same, eventually I started to doubt that. I owe you a beer! – Billy Rubina Jun 10 '18 at 1:03
• @BillyRubina Yes. That is what this exercise is for. To show that they are in fact different concepts. – user568248 Jun 10 '18 at 1:05
• But then, having partial derivatives do not imply continuity, but differentiability implies continuity? – Billy Rubina Jun 10 '18 at 1:08
• @BillyRubina That is correct. There are there are theorems to go in the other direction, imposing additional conditions of course. For examples existence of partial derivatives in a neighborhood of a point, plus them being continuous at the point, does imply differentiability at that point. – user568248 Jun 10 '18 at 1:08
• @BillyRubina Correct. Problems from lower dimensions often can be passed onto higher dimensions by defining $f(x,y,...)=h(x)$ – user568248 Jun 10 '18 at 1:36

Having partial derivatives at every point is not the same as being differentiable.

The usual formulas, i.e. in this case the quotient rule and the product rule, show that $(x,y) \mapsto \dfrac{xy}{x^2+y^2}$ has partial derivatives with respect to $x$ and $y$ at every point except the one point where the numerator and denominator are both $0.$ At that point, the existence of partial derivatives can be shown by applying the definition of differentiation: $$\left.\frac{\partial u}{\partial x}\right|_{(x,y)\,=\,(0,0)} =\lim_{\Delta x\,\to\,0} \frac{u(0+\Delta x, 0) - u(0,0)}{\Delta x} = \lim_{\Delta x\,\to\,0} \frac{{0} - 0}{\Delta x} = 0$$ and similarly for the other one.

But $u$ is not differentiable at $(0,0)$ because there is no tangent plane. Since the partial derivatives with respect to $x$ and $y$ at $(0,0)$ are both $0,$ if there is a tangent plane its equation must be $z = 0 + 0(x-0) + 0(y-0).$

However, that will not work for the following reason. Suppose one moves away from $(0,0)$ in a direction at a $45^\circ$ angle to the two coordinate axes. That is the line $x=y.$ If $x=y$, then $u(x,y) = u(x,x) = \dfrac{2x^2}{x^2+x^2} = 1.$ This is not even continuous at $(0,0)$, and if it were, if the derivative in that direction were anything but $0,$ then it would not fit the equation of the tangent plane just given.

"Has partial derivatives everywhere" means quite literally that it has partial derivatives everywhere. In other words, at every point, the partial derivatives of $u$ with respect to each variable exist. In other words, for all $(x,y)\in\mathbb{R}^2$ the limits $$u_x(x,y)=\lim_{h\to 0}\frac{u(x+h,y)-u(x,y)}{h}$$ and $$u_y(x,y)=\lim_{h\to 0}\frac{u(x,y+h)-u(x,y)}{h}$$ exist.

So, this doesn't have anything at all to do with $u$ being differentiable in your sense, at least not directly. You are correct that $u$ is not differentiable at $(0,0)$ (it is not even continuous there!), but the partial derivatives $u_x(0,0)$ and $u_y(0,0)$ still exist. Try computing them directly from the limit definitions above!