# Decide whether $f$ is differentiable in $(0,0)$ or not given its directional derivative

The function $$f: \mathbb{R}^2 \rightarrow \mathbb{R}$$ has, at the point $$(0,0)$$, for $$v \in \mathbb{R}^2\setminus \{(0,0)\}$$, the directional derivative given by $$\frac{\partial f}{\partial v} (0,0) = \frac{|v_1| v_2}{|v_1| + |v_2|} + 3v_1 - 2v_2.$$ Compute the partial derivatives at $$(0,0)$$ and determine whether $$f$$ is differentiable at $$(0,0)$$.

So I think I got the partial derivatives (by substituting with unit vectors): $$\frac{\partial f}{\partial e_1} (0,0) = \frac{|1| \cdot 0}{|1| + 0} + 3 \cdot 1 - 2 \cdot 0 = 3,$$ $$\frac{\partial f}{\partial e_2} (0,0) = \frac{|0| \cdot 1}{|0| + 1} + 3 \cdot 0 - 2 \cdot 1 = -2.$$ But I'm not so sure how one can show whether $$f$$ is differentiable in $$(0,0)$$ or not... my intuition tells me it's not differentiable.. but I feel like I don't have enough info. I was maybe thinking of showing that $$f$$ doesn't have all directional derivatives at $$(0,0)$$ but I couldn't figure it out.

• What happens if you take the directional derivative of $f$ with $v = (1, 1)$? If $f$ were differentiable, how should the directional derivative of $f$ in the $(1, 1)$ direction be related to the directional derivative in the $e_1$ and $e_2$ directions? Jun 9, 2023 at 7:37
• @Frank should the directional derivative match with e_1 or e_2? and it doesnt so its not differentiable in (0,0)? (since for v=(1,1) the d.d. is 5/2) Jun 9, 2023 at 8:39

If $$f$$ is differentiable in $$a$$ then you know that for $$v = (v_1,v_2)$$ you should have

$$\frac{\partial f}{\partial v} (a)= (\nabla f (a)) \cdot v = v_1 \frac{\partial f}{\partial e_1}(a) + v_2 \frac{\partial f}{\partial e_2}(a)$$

In particular this means that if the function $$f$$ is differentiable in $$a$$ then the function $$v \mapsto \frac{\partial f}{\partial v} (a)$$ must be a linear function, which is not the case for our function $$f$$ at $$a = (0,0)$$.

As Frank suggested, considering the vector $$v = (1,1)$$ shows the function is not differentiable. Indeed if the function is differentiable then you should have

$$\frac{\partial f}{\partial v}(0,0) = 1\frac{\partial f}{\partial e_1}(0,0) + 1 \frac{\partial f}{\partial e_2}(0,0) = 3-2 = 1.$$

But this is not the case :

$$\frac{\partial f }{\partial v} (0,0) = \frac{1}{1+1} + 3 - 2 = \frac{1}{2} + 1 = \frac{3}{2}.$$