I want to derive, w.r.t $x$, this:
$x'Ax+2y'B'x+y'Cy$
The reference says:
"Assuming $A$ positive definite, then the partial derivative is: $2(Ax+By)$."
Why the transpose $x'$ it's not in the result? Why the second term of the derivative is not $2(By)'$?
obs: The assumption, $A$ is positive definite, may not be important to this result, because my reference it's not too clear/correct sometimes.
It depends if you want your gradient as a column or row vector.
The partial derivative in $x$ is the ensemble of the partial derivative in its component $x_i$.
$$\frac{\partial}{\partial x_i} (x'Ax+2y'B'x+y'Cy) = (x'A)_i + (Ax)_i + 2(y'B')_i $$
with
$$ 2(y'B')_i = 2(By)_i$$
If $A$ is symmetric (not necessarily definite) we know that
$$ (x'A)_i + (Ax)_i = 2(Ax)_i = 2(x'A)_i $$
so in the end $$\frac{\partial}{\partial x_i} (x'Ax+2y'B'x+y'Cy) = 2(Ax+By)_i $$ $$\frac{\partial}{\partial x} (x'Ax+2y'B'x+y'Cy) = 2(Ax+By) $$