Show that the function $f(x,y) = |xy|$ is differentiable at 0, but is not of class $C^1$ in any neighborhood of 0.

Solution 1:

Your prove for differentiability is okay.

Not $C^1$:

Notice that $D_1 f$ does not exist at $(0,y)$ for any $y\ne 0$.

Differentiable:

The function $f$ is constant zero on the axes. Thus, the partial derivatives at $(0,0)$ are $0$. Hence, the candidate for $Df(0)$ is $(0,0)$. Now, for $(x,y) \ne (0,0)$ we have $$ \begin{align} |f(x,y) - f(0,0) - 0\cdot x - 0\cdot y| &= |x||y| \\ &= \min(|x|,|y|)\max(|x|,|y|) \\ &\le \min(|x|,|y|) \sqrt{x^2 + y^2}. \end{align}$$ As $\min(|x|,|y|)$ tends to $0$ for $(x,y)\to (0,0)$, we have shown that $f$ is differentiable at $(0,0)$ with $Df(0,0) = (0,0)$.