Say $\mathbf{M}$ is a symmetric, p.d. 2x2 matrix, and $\mathbf{x}$ is a 2x1 vector.
The familiar quadratic form is of course given by: $A=\mathbf{x'}\mathbf{M}\mathbf{x}$ (where $A$ is a scalar), and $$\frac{dA}{d\mathbf{x}}=2\mathbf{M}\mathbf{x}$$
Now define a new 2x1 vector $\mathbf{q}$ which is just $\mathbf{x}$ except that each element $x_i$ is raised to the power $a_i$. In other words:
$$\mathbf{q}=\begin{pmatrix} {x_1^{a_1}} \\ {x_2^{a_2}} \end{pmatrix}$$
The quadratic form $Q=\mathbf{q'}\mathbf{M}\mathbf{q}$ is then $$Q=M_{11}x_1^{2a_1}+2M_{12}x_1^{a_1}x_2^{a_2}+M_{22}x_2^{2a_2}$$
And (if I did this right):
$$\frac{dQ}{d\mathbf{x}}=\begin{pmatrix} 2a_1x_1^{a_1-1}(M_{11}x_1^{a_1}+M_{12}x_2^{a_2}) \\ 2a_2x_2^{a_2-1}(M_{12}x_1^{a_1}+M_{22}x_2^{a_2}) \end{pmatrix}$$
Which I guess can then be written more compactly using the Hadamard elementwise symbol ($\circ$): $$\frac{dQ}{d\mathbf{x}}=2\mathbf{M}\mathbf{q}\circ \mathbf{a} \circ \mathbf{q'} \mathbf{x^{\circ -1}}$$
(Where $\mathbf{a}$ is the vector of powers ($a_1,a_2$))
Is there a way to express $\frac{dQ}{d\mathbf{x}}$ as a "nicer" matrix equation without resorting to Hadamard elementwise operations?