I'm getting slightly confused about notation regarding derivatives and wanted some clarification.

Consider the function $f:\mathbb{R}^n \rightarrow \mathbb{R}$.

I know that $Df = (D_1f, D_2f, ..., D_nf) =$ gradient of $f$.

I was just wondering what this notation $D^kf$ for some $k \geq 2$ means. Is it a vector or a matrix?

This is the definition I'm referring to: $$ \beta=\left(\beta_1,\dots,\beta_n\right),\beta_i=\text{integer}\geqslant 0, \text{with} \left\vert\beta\right\vert=\sum\beta_i, \text{is a $multi$-$index$}; \text{we define} \\ D^\beta u = \frac{\partial^{\left\vert\beta\right\vert}u}{\partial x_1^{\beta_1} \dots \partial x_n^{\beta_n}} $$

  • $\begingroup$ For $k=2$ it could be the Hessian matrix but who knows. $\endgroup$
    – Kurt G.
    Commented Jan 18, 2023 at 14:19
  • $\begingroup$ So for $k = 3$ would $D^3f$ be a matrix with entries $D_{ijk}$ for each $i,j,k = 1,...,n?$. $\endgroup$ Commented Jan 18, 2023 at 14:38
  • $\begingroup$ I object to elaborate since such notational questions with zero context allow almost every answer. Just one thing: if we assume the entries have three indices is that a matrix ? Don't think so. $\endgroup$
    – Kurt G.
    Commented Jan 18, 2023 at 14:43
  • $\begingroup$ Aha. $\boldsymbol{\beta}$ is a multi-index (a vector of indices), not a $k\ge 2\,.$ What exactly is unclear now? $\endgroup$
    – Kurt G.
    Commented Jan 18, 2023 at 14:54

1 Answer 1


I would read it this way:

Let $\vec{k} \in \mathbb{N}_0^n$ be a vector of indices. $$ D^\vec{k} f = \frac{\partial^{\left\vert \vec{k}\right\vert}f}{\partial x_1^{k_1} \cdots \partial x_n^{k_n}} = \frac{\partial^{\sum_{i=1}^n k_i}f}{\prod_{i=1}^n \partial x_i^{k_i}} $$ So the result will be the function $f$ derived $k_i$ times by $x_i$ for all $1 \le i \le n$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .