 Solution

Write $\dot{x}=dx/du$, $\ddot{x}=d^{2}x/d{u}^2$, $A_v=dA/dv$, $A_{vv}=d^{2}A/d{v}^2$, and $y_w=dy/dw$, $y_{ww}=d^{2}y/d{w}^2$. Put $S=A+A^T$. The product rule and chain rule giveHere $w^{\prime },w^{\prime \prime }$ are derivatives with respect to $v$. In the first derivative, $x^{T}A\dot{x}={\dot{x}}^T{A}^{T}x$ because the expression is scalar. Likewise ${\dot{x}}^{T}S\dot{x}=2{\dot{x}}^{T}A\dot{x}$ if desired. This differentiation of a parameter-dependent quadratic form does not assume that $A$ is symmetric. The positive linear term $y^{T}b$ contributes the positive chain-rule terms in the $v$ derivatives; it is independent of $u$.