Iterated derivative

Definition

Let $U \subseteq \mathbb{R}^n$ be open and let $f \colon U \to \mathbb{R}^m$. For an integer $k \ge 0$, let $L^k(\mathbb{R}^n; \mathbb{R}^m)$ be the space of $k$-linear maps from $(\mathbb{R}^n)^k$ to $\mathbb{R}^m$, with $L^0(\mathbb{R}^n; \mathbb{R}^m) = \mathbb{R}^m$. The iterated derivatives are defined by recursion. First, $D^0 f = f$. Suppose that $D^k f \colon U \to L^k(\mathbb{R}^n; \mathbb{R}^m)$ is defined on $U$ and is differentiable on U. Then $D^{k+1} f(a)$, for $a \in U$, is the $(k+1)$-linear map $$D^{k+1} f(a)(v_1, v_2, \dots, v_{k+1}) = \big(D(D^k f)(a)(v_1)\big)(v_2, \dots, v_{k+1}).$$ Here $D(D^k f)(a)$ is the Fréchet derivative of $D^k f$ at $a$, a linear map from $\mathbb{R}^n$ to $L^k(\mathbb{R}^n; \mathbb{R}^m)$.

Remarks

For $k = 1$, $D^1 f = Df$. For $k = 2$, $D^2 f(a)(u, v)$ is the derivative of $x \mapsto Df(x)(v)$ at $a$ in the direction $u$. On basis vectors, $D^2 f(a)(e_i, e_j) = \dfrac{\partial}{\partial x_i}\left(\dfrac{\partial f}{\partial x_j}\right)(a)$. In general, $D^k f(a)(e_{i_1}, \dots, e_{i_k})$ is a partial derivative of order k. For a real-valued $f$, the matrix of the bilinear map $D^2 f(a)$ is the Hessian matrix. If $f$ is of class $C^k$, then $D^k f(a)$ is a symmetric $k$-linear map, by the symmetry of second derivatives.

Explanations

L1 human

The first derivative takes one direction and gives a rate of change. The second derivative takes two directions $u$ and $v$. It tells how the rate of change in direction $v$ itself changes as you move in direction $u$.