Consider the expression $e^{-\langle x,t\rangle}$, where $x$ and $t$ are both $m$-dimensional vectors.
I am interested in evaluating the following expression
$$ \frac{d^n}{dx^n}e^{-\langle x,t\rangle} $$
For the first derivative, I can clearly see we have $$ \frac{d}{dx}e^{-\langle x,t\rangle} = -xe^{-\langle x,t\rangle} $$ But then how to proceed? Often when I've been faced with calculating higer-order derivatives of scalar functions I see something like $\frac{d^2}{dxdx^{T}}$, but if I continue this alternating of transposes I'll either get the derivative sequentially evaluating between a vector and a scalar, or between a scalar and a matrix, and that doesn't feel right.
It may help to use the index notation, $$ f(\vec x) = \exp(-x_i t_i) $$ Then $$ \frac{d f}{dx_j} = -\exp(-x_i t_i) t_i \delta_{ij} = -\exp(-x_i t_i)t_j, $$ where $\delta_{ij} = 1$ if $i = j$ and zero otherwise. This can be written compactly as $\nabla f = -\exp(-\langle x,t \rangle) t$. Taking a second derivative $$ \frac{d^2 f}{dx_j dx_k} = \exp(-x_i t_i)t_j t_k. $$ This can be written compactly as $\nabla^2 f = \exp(-\langle x,t \rangle) t t^T$. Each derivative will add an extra dimension. Thus a third (or higher) derivative will be a tensor.