# Every continuous function has directional derivative? [closed]

I understand a continuous function may not be differentiable. But does every continuous function have directional derivative at every point? Thanks!

• A continuous not differentiable function on $\mathbb R$ will be a counterexample. Commented Sep 21, 2020 at 17:29
• @OlivierMoschetta That is not correct, one needs some sort of uniformity. Commented Sep 21, 2020 at 17:43
• The function $f(x) = \sqrt{|x|}$ is continuous but does not have a directional derivative at $x=0$. Commented Sep 21, 2020 at 17:44
• Hint: In one dimension, the derivative is a directional derivative. Commented Sep 21, 2020 at 17:52
• In fact, there exist continuous functions from ${\mathbb R}^n$ to $\mathbb R$ such that, at EACH point, NONE of the directional derivatives exist at that point. An example for $n=2$ (which can be easily adapted for larger values of $n)$ is given in Theorem 4 on p. 973 of Crinkly curves and choppy surfaces by Felix Adalbert Behrend. (continued) Commented Sep 21, 2020 at 18:04

No. Consider the function $$f(x,y) = e^{-\sqrt{x^2+y^2}}$$. Then $$f$$ is continuous everywhere, but $$f(0,0)$$ has no directional derivative at $$(0,0)$$. I'll let you prove this rigorously on your own, but this should be clear from the plot below; the graph is obviously not differentiable at the "tip".
There is no need to distinguish between one and more directions. If a continuous function like $$x\longmapsto |x|$$ isn't differentiable at $$x=0,$$ then there is no directional derivative in $$x$$-direction.
The contrary is the case: every differentiation is always a directional differentiation. We have that $$f$$ is differentiable at $$x_0$$ if there is a linear function $$\mathbf{J}$$ and a remainder function $$\mathbf{r}$$ such that $$$$\mathbf{f(x_{0}+v)=f(x_{0})+J(v)+r(v)}$$$$ where $$\mathbf{v}$$ is the direction where we consider a change of slope. Partial derivatives are along the coordinate axis, the total derivative is a linear combination of partial derivatives, but all are in some direction $$\mathbf{v}$$.