# Directional Derivatives of Functions from Rn to Rm and Continuity

Recall that if $f : S \to \mathbb{R}$ where $S \subseteq \mathbb{R}$ is open, then $f$ being differentiable at a point $x_0 \in S$ implies that $f$ is continuous at $x_0$ (the converse being false).

We would like to develop a derivative definition for functions $\mathbf{f} : S \to \mathbb{R}^m$ where $S \subseteq \mathbb{R}^n$ is open that preserves this nice property. Recall from the Directional Derivatives of Functions from Rn to Rm page that the directional derivative of $\mathbf{f}$ at $\mathbf{c}$ in the direction of the vector $\mathbf{u} \in \mathbb{R}^n$ is defined as the following limit provided that it exists:

(1)One might ask, does the existence of all directional derivatives of a function $\mathbf{f}$ at $\mathbf{c}$ imply that $\mathbf{f}$ is continuous at $\mathbf{c}$? The answer is sadly NO. For example, consider the following function $f : \mathbb{R}^2 \to \mathbb{R}$ defined by:

(2)Let's compute the directional derivatives of this function at the origin $\mathbf{c} = (0, 0)$. Let $\mathbf{u} = (u_1, u_2) \in \mathbb{R}^2$. Then:

(3)Notice that $f'(\mathbf{c}, \mathbf{u})$ exists whenever $u_2 \neq 0$. Suppose now that $u_2 = 0$. I.e., let's compute the directional derivative of $f$ at $\mathbf{c}$ in the direction of $\mathbf{u} = (1, 0)$ (the partial derivative of $f$ with respect to the first variable). Then:

(4)So indeed all directional derivatives of $f$ at $\mathbf{c}$ exist. However, we claim that $f$ is discontinuous at $\mathbf{c}$. We first give a graph of $f$ below to visualize the discontinuity at $\mathbf{c} = (0, 0)$:

Now, along the line $y = x^2$ we have that:

(5)But $\displaystyle{f(0, 0) = 0 \neq \frac{1}{2}}$ so $f$ is discontinuous at $\mathbf{c} = (0, 0)$ despite having all of its directional derivatives existing at $\mathbf{c}$.