calculate derivative $\frac{1}{2}(x+\alpha p)^T A(x+\alpha p)-b^T(x+\alpha p)$ about $\alpha$

Question

Suppose $A$ is a $n\times n$ matrix, $x,p$ are $n$ dimensional vector. Calculate the derivative $\frac{1}{2}(x+\alpha p)^T A(x+\alpha p)-b^T(x+\alpha p)$ about $\alpha$.

Answer 1

$$ \frac{1}{2}(x+\alpha p)^T A(x+\alpha p)-b^T(x+\alpha p) = \frac{1}{2}p^T Ap \alpha^2 +(p^TAx-b^T p) \alpha+\frac{1}{2}x^TAx-b^Tx $$

$$ \frac{\partial }{\partial \alpha}\left[\frac{1}{2}(x+\alpha p)^T A(x+\alpha p)-b^T(x+\alpha p)\right]=p^T A p \alpha + (p^TAx-b^T p) $$