I honestly cannot find something similar to this function online - every other case I have found (and which seems to be very standard) is $x^T$Ax.
A question my professor asked involved finding the derivative of f($\mathbf x$) = A($\mathbf x$)$\bullet$$\mathbf x$, f: $ℝ^n$ -> ℝ.
For context: he wants us to use Lagrange multipliers to show that if $\mathbf x$ is a local extremum of f subject to constraint ||$\mathbf x$|| = 1, then there exists $\lambda \in$ ℝ s.t. A$\mathbf x$ = $\lambda$$\mathbf x$
I want to give an honest try at the proof before I ask anyone for help, so I just provided it for context in case it makes a difference. I just can't for the life of me figure out the basic derivative! My gut is saying that it's a simple product rule application, but I'm not sure.