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

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    $\begingroup$ A continuous not differentiable function on $\mathbb R$ will be a counterexample. $\endgroup$ Commented Sep 21, 2020 at 17:29
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    $\begingroup$ @OlivierMoschetta That is not correct, one needs some sort of uniformity. $\endgroup$
    – copper.hat
    Commented Sep 21, 2020 at 17:43
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    $\begingroup$ The function $f(x) = \sqrt{|x|}$ is continuous but does not have a directional derivative at $x=0$. $\endgroup$
    – copper.hat
    Commented Sep 21, 2020 at 17:44
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    $\begingroup$ Hint: In one dimension, the derivative is a directional derivative. $\endgroup$
    – John Douma
    Commented Sep 21, 2020 at 17:52
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    $\begingroup$ 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) $\endgroup$ Commented Sep 21, 2020 at 18:04

2 Answers 2

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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". enter image description here

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  • $\begingroup$ Approaching the origin from any angle, the slope has a limit. What is that value if not a directional derivative? $\endgroup$
    – Tavin
    Commented Jul 16, 2023 at 21:20
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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 \begin{equation}\mathbf{f(x_{0}+v)=f(x_{0})+J(v)+r(v)} \end{equation} 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 $\mathbf{v}$.

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