I am trying to compute the following vector-by-vector derivative
$$ \frac{\text{d}}{\text{d}\mathbf{x}}\left(\mathbf{A}\mathbf{x}\mathbf{x}^T\mathbf{A}\mathbf{\dot{x}}\right), $$
where $\mathbf{x}$ and $\mathbf{\dot{x}}$ are $n \times 1$ column vectors, $\mathbf{A}$ is a constant $n \times n$ matrix, and $\dfrac{\text{d}\mathbf{\dot{x}}}{\text{d}\mathbf{x}}$ is a known $n \times n$ matrix.
I try to solve this by switching to index notation: $$ \begin{aligned} \frac{\text{d}}{\text{d}x_m}\left(A_{ij} x_j x_k A_{kl} \dot{x}_l\right) &= A_{ij} \delta_{jm} x_k A_{kl} \dot{x}_l + A_{ij} x_j \delta_{km} A_{kl} \dot{x}_l + A_{ij} x_j x_k A_{kl} \frac{\text{d}\dot{x}_l}{\text{d}x_m} \\ &\Rightarrow \mathbf{A} \left(\mathbf{x}^T \mathbf{A} \mathbf{\dot{x}}\right) + \mathbf{A} \mathbf{x} \,\left(\mathbf{A} \mathbf{\dot{x}}\right)^T + \mathbf{A} \mathbf{x} \mathbf{x}^T \mathbf{A} \dfrac{\text{d}\mathbf{\dot{x}}}{\text{d}\mathbf{x}} \end{aligned} $$
Is this correct? I don't have much experience with this kind of thing. Thank you.
Indeed I think it is correct. You can use differential to avoid index notations.
Denote first $$ \phi(\mathbf{x}) = \mathbf{A}\mathbf{x} (\mathbf{x}^T\mathbf{A}\mathbf{\dot{x}}) $$ where the term in bracket is a scalar $ s =\mathbf{x}^T (\mathbf{A}\mathbf{\dot{x}}) = (\mathbf{A}\mathbf{\dot{x}})^T \mathbf{x} $
From here, it follows $d\phi = s\cdot d(\mathbf{Ax}) + (\mathbf{Ax}) \cdot ds = s\mathbf{A} d\mathbf{x} + (\mathbf{Ax}) \cdot ds $
It remains to compute $ds = (\mathbf{A}\mathbf{\dot{x}})^T d\mathbf{x} + \mathbf{x}^T \mathbf{A} d\mathbf{\dot{x}} $.
Using $d\mathbf{\dot{x}} = \dfrac{\text{d}\mathbf{\dot{x}}}{\text{d}\mathbf{x}} d\mathbf{x}$, and reorgainzing everything, we end up with
$$ d\phi = \left[ s\mathbf{A} + (\mathbf{Ax}) (\mathbf{A}\mathbf{\dot{x}})^T + \mathbf{Ax} \mathbf{x}^T \mathbf{A} \dfrac{\text{d}\mathbf{\dot{x}}}{\text{d}\mathbf{x}} \right] d\mathbf{x} $$ from which you easily obtain the vector derivative.