Justifying $\frac{d A^Tx^TAx}{dx} = 2A^TAx$

Question

When deriving the normal equations, I have trouble justifying why the derivative of $A^Tx^TAx$ is actually $2A^TAx$.

I know that the derivative of $x^TAx$ wrt $x$ is $2Ax$ when $A$ is symmetric.

Is it correct to say that because $A^TA$ is symmetric, one can simply state that $\frac{d A^Tx^TAx}{dx} = 2A^TAx$?

Is it that obvious or am I missing something?

Answer 1

$ F = A^T (x^T A x) $

$\nabla F = A^T \nabla (x^T A x) = A^T (A + A^T ) x $