Hessian matrix of a quadratic form

So let's compute the first derivative, by definition we need to find $f'(x)\colon\mathbb R^n \to \mathbb R^n$ such that $$ f(x+h) = f(x) + f'(x)h + o(h), \qquad h \to 0 $$ We have \begin{align*} f(x+h) &= (x+h)^tA(x+h)\\ &= x^tAx + h^tAx + x^tAh + h^tAh\\ &= f(x) + x^t(A + A^t)h + h^tAh \end{align*} As $|h^tAh|\le \|A\||h|^2 = o(h)$, we have $f'(x) = x^t(A + A^t)$ for each $x \in \mathbb R^n$. Now compute $f''$, we have \begin{align*} f'(x+h) &= x^t(A + A^t) + h^t(A + A^t)\\ &= f(x) + h^t(A + A^t) \end{align*} So $f''(x) = A + A^t$.

Intuitively, the gradient and Hessian of $f$ satisfy \begin{equation} f(x + \Delta x) \approx f(x) + \nabla f(x)^T \Delta x + \frac12 \Delta x^T Hf(x) \Delta x \end{equation} and the Hessian is symmetric.

In this problem, \begin{align*} f(x + \Delta x) &= (x + \Delta x)^T A (x + \Delta x) \\ &= x^T A x + \Delta x^T A x + x^T A \Delta x + \Delta x^T A \Delta x \\ &= x^T A x + \Delta x^T(A + A^T)x + \frac12 \Delta x(A + A^T) \Delta x. \end{align*}

Comparing this with the approximate equality above, we see that $\nabla f(x) = (A + A^T) x$ and $Hf(x) = A + A^T$.

Write explicitly $$f(x)=\sum_{i,j}(\text{2nd degree monomials})$$ The hessian is the matrix $$H=(\partial_i\partial_jf(x)).$$