Derivative vanishes at a local extremum

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}$, and let $a \in U$ be a local extremum of $f$. If the partial derivative $\partial f / \partial x_i (a)$ exists for some $i$, then $\partial f / \partial x_i (a) = 0$. In particular, if $f$ is differentiable at $a$, then $\nabla f(a) = 0$ and the derivative $Df(a)$ is the zero linear map.

Remarks

A point $a \in U$ where $f$ is differentiable and $\nabla f(a) = 0$ is a critical point, or stationary point, of $f$. The proof applies the one-variable theorem to $t \mapsto f(a + t e_i)$, where $e_i$ is the $i$-th standard basis vector. This function has a local extremum at $t = 0$. The converse is false. For $f(x, y) = x^2 - y^2$, $\nabla f(0, 0) = 0$, but $(0, 0)$ is a saddle point, not an extremum. The hypothesis that $U$ is open matters. On $[0, 1]^2$, $f(x, y) = x + y$ has a maximum at $(1, 1)$, where $\nabla f = (1, 1)$.

Explanations

L1 human

At the top of a smooth hill the ground is flat in every direction. So all partial derivatives are zero there, and the tangent plane to the graph is horizontal. This turns the search for extrema into the solution of the equations $\nabla f = 0$.