I am studying optimal control (for an introductory course) and we have to know the derivatives of quadratic forms.
Out teacher said that the following are true: $$\frac{\partial Ax}{\partial x} = A^T $$ $$ \frac{\partial x^TAx}{\partial x} = (A^T+A)x $$
from what i gathered (using this source: https://www.kamperh.com/notes/kamper_matrixcalculus13.pdf ) the derivative of a scalar function is defined as: $$\frac{\partial f(x)}{\partial x} = \begin{bmatrix} \frac{\partial f}{\partial x_1} \\ \frac{\partial f}{\partial x_2} \\ ... \\ \frac{\partial f}{\partial x_n} \end{bmatrix} $$
however, i have been taught the derivative so far as (also presented here http://michael.orlitzky.com/articles/the_derivative_of_a_quadratic_form.xhtml) : $$\frac{\partial f(x)}{\partial x} = \begin{bmatrix} \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, ... , \frac{\partial f}{\partial x_n} \end{bmatrix} $$
This way of thinking about the derivative is combatible witht he taylor series: $$ f(x+h) = f(x) + \frac{\partial f(x)}{\partial x} \cdot h $$
And the previous derivatives now become (for combatible dimensions. See also the second link where the folling are prooved) : $$\frac{\partial Ax}{\partial x} = A $$ $$ \frac{\partial x^TAx}{\partial x} = x^T(A^T+A) $$
Is someone wrong? If that is not the case, what is the motivation (and advantages ?) for the second definition?
