# Derivative notation assistance

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

• For $k=2$ it could be the Hessian matrix but who knows. Commented Jan 18, 2023 at 14:19
• So for $k = 3$ would $D^3f$ be a matrix with entries $D_{ijk}$ for each $i,j,k = 1,...,n?$. Commented Jan 18, 2023 at 14:38
• 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. Commented Jan 18, 2023 at 14:43
• Aha. $\boldsymbol{\beta}$ is a multi-index (a vector of indices), not a $k\ge 2\,.$ What exactly is unclear now? Commented Jan 18, 2023 at 14:54

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 $$1 \le i \le n$$.