Counterexample of linearity of the derivative

Question

I found that the directional derviative $D_xf(0,0) = \sqrt{r^2+s^2} \cdot g \left ( \dfrac{r}{\sqrt{r^2+s^2}}, \dfrac{s}{\sqrt{r^2+s^2}} \right )$ for $x = (r,s)$

and I am then asked to show

I can't find a counter example

Answer 1

I think the following is valid: you can write $f(r,\theta) = rg(\theta)$ in polar coordinates and compute the gradient $\nabla f = g(\theta) \hat r(\theta) + g'(\theta)\hat\theta(\theta)$. So for general $g$ the gradient isn't well defined at the origin - it depends on the path travelled through the origin. I think this is still true given the conditions imposed on $g$ in your question. (With those conditions, $g$ can be written as $g(\theta) = \sum_{n=2}^{\infty} g_n \sin n\theta$, and so it still has $\theta$-dependence that makes it undefined at the origin.)

You can still (informally) define the directional derivative as $D_xf=x\cdot \nabla f$ because that exists as a limit at the origin (as you travel through the origin, $\hat r(\theta)$ and $g(\theta)$ both flip signs so the limit ends up being the same from both directions). But the $\theta$-dependence of $\nabla f$ makes it so that $D_xf$ isn't linear in $x$ the way it should be.