Showing that a function is not differentiable

Solution 1:

Use directional derivatives. Note that the limit of $$\frac{f(h,h)-f(0,0)}{h\sqrt{2}}=\frac{|h|}{h\sqrt{2}}$$ as $h\to 0^+$ is different than that as $h\to 0^-$.

Solution 2:

Note: My previous answer was incorrect, thanks to @Lubin for catching that.

Simplify your life and show that $\phi(x) = f(x,x) = |x|$ is not differentiable at $0$. It will follow from this that $f$ is not differentiable at $0$.

Look at the definition of differentiability for this case, which is that $\lim_{h \to 0, h \neq 0} \frac{\phi(h)-\phi(0)}{h} $ exists. We have $\phi(0) = 0$, and $\phi(h) = |h|$, so we are looking at the limit of $h \mapsto \mathbb{sign}(h)$ as $h \to 0$.

If you choose $h_n = \frac{(-1)^n}{n}$, it is easy to see that $\frac{\phi(h_n)-\phi(0)}{h_n} = (-1)^n$, hence it has no limit. It follows that $f$ is not differentiable at $0$.

Solution 3:

In agreement with @Cameron, I would show the nondifferentiability of $f(x,y)$ at $(0,0)$ in the following way. Intersect the graph with a vertical plane through the $z$-axis, say given by $y=\lambda x$. The intersection is given by $z=|\lambda|^{1/2}\cdot|x|$. So above each of the four quadrants of the $x,y$-plane, the graph consists of strings stretched out from the origin at varying angles. In particular, the “diagonal” plane $y=x$ intersects the graph in a V-shaped figure exactly like the familiar graph of absolute value in one-variable calculus.