Does the expression of a directional derivative in terms of normal derivatives require continuity?

I read on wikipedia that, if $$\mathbf{v}=\langle v_1,v_2\rangle$$ is a unit vector in $$\mathbb{R}^2$$, then the directional derivative $$\nabla_{\mathbf{v}}f(x,y)=v_1\frac{\partial f}{\partial x}+v_2\frac{\partial f}{\partial y}$$ if the partial deriavtives are continuous at $$(x,y)$$. However, nowhere else states that it is required that the partial derivatives be continuous, only that (obviously) they must exist. Can someone clarify whether or not it is true that the derivatives must be continuous, and if it is true, can you please explain why?

Wikipeida article that says the derivatives must be continuous: https://en.wikipedia.org/wiki/Derivative#Directional_derivatives

Other places that don't: http://mathonline.wikidot.com/higher-order-directional-derivatives

http://mathworld.wolfram.com/DirectionalDerivative.html

• Existence of partial derivatives does not imply existence of directional derivative in any arbitrary direction, at a given point (x, y). Continuity of the partials is a generous way to guarantee that they exist in a neighborhood of (x, y), in which case directional derivatives indeed exist in all directions. – John Jiang Dec 29 '18 at 21:53
• This may be of interest math.harvard.edu/archive/21a_fall_09/exhibits/bitch – D.R. Dec 29 '18 at 22:01

The directional derivative of $$f$$, at $$x$$, in direction $$u$$ is defined like so:
$$D_u f(x) = \lim_{t \to 0} \frac{f(x + tu) - f(x)}{t}.$$
This can exist for very non-smooth functions. For example, when $$f$$ is the indicator function for the $$x$$-axis (i.e. takes the value $$1$$ on $$\mathbb{R} \times \{ 0 \}$$ and $$0$$ elsewhere), then the directional derivative at $$0$$ exists in the directions $$(1, 0)$$ and $$(-1, 0)$$ (and is equal to $$0$$). Note that the $$y$$-partial derivative doesn't exist in this case, and the dot product formula is nonsense.
If $$f$$ is differentiable in the multivariate sense (one sufficient condition is that the partial derivatives exist in a neighbourhood of the point and are continuous at the point), then the dot product formula holds. In fact, the existence of a vector $$v$$ such that, for all direction vectors $$u$$,
$$v \cdot u = D_u f(x)$$
is practically the definition of $$f$$ being differentiable at the point $$x$$. Such a vector must necessarily be $$\nabla f (x)$$.