Hessian of $f(x) = e^{u^Tx}$ for some $u\in \mathbb{R^d}$

Question

I'm seeing the hessian is $uu^Te^{u^Tx}$. From my understanding since this is a single variable function, we can basically treat $u^T$ as a constant and the hessian is equivalent to the second derivative of f(x) with respect to x. My question is why is the solution $uu^Te^{u^Tx}$ and not $u^Tu^Te^{u^Tx}$. Is this because the vector dimensions for $u^Tu^T$ don't work for multiplication? - If that's the case, why can we just swap $u^Tu^T$ with $uu^T$?

Answer 1

Let $x=(x_1,....,x_d)$ and $u=(u_1,...,u_d).$ Then:

$$f(x)=\prod_{j=1}^d e^{x_ju_j}.$$

This gives

$$f_{x_k}=u_kf(x).$$

and

$$f_{x_ix_k}=u_iu_kf(x).$$

Thus the Hessian is given by

$$uu^Tf(x)=uu^Te^{u^Tx}.$$