TITLE: Why is $\mathbf{d}^{t} H \mathbf{d}$ the second directional derivative?
QUESTION [5 upvotes]: Given a function $f: \mathbb{R}^m \to \mathbb{R}$, why does $\mathbf{d}^{t} H \mathbf{d}$ give the second derivative to $f$ along the unit vector $\mathbf{d}$? Here $H$ is the Hessian, so that $H_{ij} = \frac{\partial f}{\partial x_i \partial x_j}$. I suppose I don't have the right definition to work with, so any intuition would be helpful, as well as formal proof. (e.g. why this works in the second order taylor series expansion)

REPLY [2 votes]: Consider the function of one variables given by $g(t)=f(x+td)$. You want to compute $g''(0)$. So you need to use the chain rule. You get $g'(t)=\sum_i\frac{\partial f}{\partial x_i}(x+td)d_i$ and 
$g''(t)=\sum_i\sum_j\frac{\partial^2 f}{\partial x_i\partial x_j}(x+td)d_i d_j$. So
$g''(0)=\sum_i\sum_j\frac{\partial^2 f}{\partial x_i\partial x_j}(x)d_i d_j= d^tH(x)d$.