# Is $f(x,y)=(xy)^{2/3}$ differentiable at $(0,0)$?

I looked at the graph of $f(x,y)=(xy)^{2/3}$ at the origin, and no matter how far I zoom in it has four corners along $x=\pm y$, so it doesn't look like it is differentiable as $z=0$ doesn't look like a tangent plane.

I also found that its partial derivatives are not continuous at $(0,0)$, e.g. $$\lim_{(ky^2,y) \to (0,0)} \frac{\partial f}{\partial x} = \lim_{(ky^2,y) \to (0,0)} \frac{2}{3}x^{-\frac{1}{3}}y^{\frac{2}{3}} = \frac{2}{3}k^{-\frac{1}{3}}$$ (although $\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0) = 0$)

I also tried to prove/disprove its differentiability by definition, and for it to be differentiable I need $$\lim_{(x,y) \to (0,0)} \frac{f(x,y)-0}{\sqrt{x^2+y^2}} =0$$ but then I got stuck. Am I even on the right track?

[UPDATE] I need to do the same thing for $f(x,y)=(xy)^{\frac{1}{3}}$... I think the answer is that $(xy)^{2/3}$ is differentiable but $(xy)^{1/3}$ is not, according to my professor but I'm not sure

• I just wonder if $x<0$, $y>0$, how can we define $f(x,y)$ ? – Canardini Dec 10 '16 at 7:48
• Only use real roots – ladiesman Dec 10 '16 at 8:09
• How does $\frac{\partial f}{\partial x}(0,0)=0$? Isn't that undefined? – INQUISITOR Jan 27 '20 at 0:16

Neither functions have continuous partials at $(0,0)$, but as you said $f(x,y)=(xy)^{\frac{2}{3}}$ is surprisingly differentiable at $(0,0)$ but $f(x,y)=(xy)^{\frac{1}{3}}$ is not.

For $f(x,y)=(xy)^{\frac{2}{3}}$, continuing with where you left off: $$\lim_{(x,y) \to (0,0)} \frac{f(x,y) - 0}{\sqrt{x^2+y^2}} = \lim_{(x,y) \to (0,0)} \frac{((xy)^2)^{\frac{1}{3}}}{\sqrt{x^2+y^2}} \le \lim_{(x,y) \to (0,0)} \frac{((x^2+y^2)^2)^{\frac{1}{3}}}{(x^2+y^2)^{\frac{1}{2}}} = \lim_{(x,y) \to (0,0)} (x^2+y^2)^\frac{1}{6}$$ where we used the inequality $xy \le x^2 + y^2$. The last expression approaches $0$ no matter how $(x,y)$ approaches $(0,0)$ (in polar coordinates $r^{\frac{1}{3}} \to 0$ as $r \to 0$), so we conclude $f$ is indeed differentiable at $(0,0)$ by definition (no matter how crazy and unconvincing the graph looks).

For $f(x,y)=(xy)^{\frac{1}{3}}$, which also has $z=0$ as the candidate tangent plane, repeat the same procedure: $$\lim_{(x,y) \to (0,0)} \frac{f(x,y) - 0}{\sqrt{x^2+y^2}} = \lim_{(x,y) \to (0,0)} \frac{(xy)^{\frac{1}{3}}}{\sqrt{x^2+y^2}}$$ here we have a smaller numerator and the same trick doesn't work as we get something like $\lim_{(x,y) \to (0,0)} (x^2+y^2)^{-\frac{1}{6}}$ which blows up as we approach the origin. In fact the limit does not exist: $$\lim_{(x,0) \to (0,0)} \frac{(xy)^{\frac{1}{3}}}{\sqrt{x^2+y^2}} = 0$$ however $$\lim_{(x,x) \to (0^+,0^+)} \frac{(xy)^{\frac{1}{3}}}{\sqrt{x^2+y^2}} = \lim_{(x,x) \to (0^+,0^+)} \frac{1}{\sqrt{2}} x^{-\frac{1}{3}} = \infty \\ \lim_{(x,x) \to (0^-,0^-)} \frac{(xy)^{\frac{1}{3}}}{\sqrt{x^2+y^2}} = \lim_{(x,x) \to (0^-,0^-)} -\frac{1}{\sqrt{2}} x^{-\frac{1}{3}} = -\infty$$ so we conclude $f(x,y)=(xy)^{\frac{1}{3}}$ is not differentiable at $(0,0)$, although its graph doesn't look much crazier than that of $(xy)^{\frac{2}{3}}$ .

The lessons are:

1. Continuous partials are not necessary (although sufficient) for differentiability;
2. Graphs can occasionally be counter-intuitive!

You have two options: 1) Go to polar coordinates (x=rcos$\alpha$ and y=rsin$\alpha$ where $r=\sqrt{x^2+y^2}$). And 2) This function converges to zero in $(0,0)$, so you need to find a sequence of $(u_k,v_k)$ such that $(u_k,v_k)$ $\to (0,0)$ but $f(u_k,v_k)\neq0$.