# How can we see that $f(x,y)=\sqrt{|xy|}$ has partial derivatives at $(0,0)$ from its graph?

How can we see that $f(x,y)=\sqrt{|xy|}$ is not differentiable at $(0,0)$ from its graph but why its partial derivatives exist? We can prove that $f(x,y)$ is not differentiable at $(0,0)$ from the definition of differentiability for multivariate functions: $$\triangle f-f(x_0,y_0)-f_x\cdot\triangle x-f_y\cdot\triangle y=\epsilon_1\cdot \triangle x+\epsilon_2\cdot\triangle y$$ by finding that epsilons that don't go to $0$.

I guess geometrically we can see that there's a "sharp" downward decline at $(0,0)$. But what is the geometrical explanation for the existence of partial derivatives?

Because if we slice the function graph over $x$ or $y$ we still see the sharp downward decline.

Here's a link to the interactive graph: https://ggbm.at/ukVg8PaR

## 1 Answer

If we want to compute the partial derivatives at the origin, the only thing that matters is how the function behaves on the coordinate axes (close to the origin), and there it is identically $0$.

Note that (one of the) partial derivatives fail to exist at all other points on the axes, because of the sharp cusp you describe, but at the origin itself, this doesn't matter.