Derivative gives the directional derivatives

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}^m$, and let $a \in U$. Suppose that $f$ is differentiable at $a$ with derivative $Df(a)$. Then for each $v \in \mathbb{R}^n$, $f$ has the directional derivative $Df(a)(v)$ at $a$ in the direction $v$: $$\lim_{t \to 0} \frac{f(a + t v) - f(a)}{t} = Df(a)(v).$$

Remarks

For $v = e_i$ this gives $\dfrac{\partial f}{\partial x_i}(a) = Df(a)(e_i)$. So the partial derivatives of the components of $f$ are the entries of the matrix of $Df(a)$. For a real-valued $f$, it gives $D_v f(a) = \nabla f(a) \cdot v$, where $\nabla f(a)$ is the gradient. In particular, $D_v f(a)$ is linear in $v$. A function whose directional derivatives are not linear in $v$ is not differentiable.

Explanations

L1 human

Once you know the linear map $Df(a)$, you know the slope of $f$ in every direction: feed the direction into the map. For $f(x, y) = x^2 + y^2$ at $(1, 1)$, $Df(1, 1)(v) = 2 v_1 + 2 v_2$. So the slope in the direction $(1, -1)$ is $0$.