---
tag: 0A6PMR20N4
kind: statement
role: proposition
name: Derivative gives the directional derivatives
msc: [26B05]
uses: [0135XZZQ8M, 05EGFBZTGC, 09EGC6F8PT]
sources:
  - url: https://en.wikipedia.org/wiki/Directional_derivative
    locator: For differentiable functions
  - url: https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative
    locator: Finite dimensions
---

## 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|0135XZZQ8M]] at $a$ with derivative $Df(a)$.
Then for each $v \in \mathbb{R}^n$, $f$ [[has the directional derivative|05EGFBZTGC]] $Df(a)(v)$ 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|09EGC6F8PT]] 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|0ABK9V9Y64]].
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 $0$.