How to take the gradient of the quadratic form? It's stated that the gradient of:
$$\frac{1}{2}x^TAx - b^Tx +c$$
is 
$$\frac{1}{2}A^Tx + \frac{1}{2}Ax - b$$
How do you grind out this equation? Or specifically, how do you get from $x^TAx$ to $A^Tx + Ax$?
 A: There is another way to calculate the most complex one, $\frac{\partial}{\partial \theta_k} \mathbf{x}^T A \mathbf{x}$. It only requires nothing but partial derivative of a variable instead of a vector.
This answer is for those who are not very familiar with partial derivative and chain rule for vectors, for example, me. Therefore, although it seems long, it is actually because I write down all the details. :)
Firstly, expanding the quadratic form yields:
$$
\begin{align}
f = \frac{\partial}{\partial \theta_k} \mathbf{x}^T A \mathbf{x} &= \frac{\partial}{\partial \theta_k} \sum_{i=1}^N \sum_{j=1}^N a_{ij}\frac{\partial}{\partial \theta_k}(\mathbf{x}_i \mathbf{x}_j)
\end{align}
$$
Since 
$$
\frac{\partial}{\partial \theta_k}(\mathbf{x}_i \mathbf{x}_j) = 
\begin{cases}
\mathbf{x}_j, && \text{if } k = i \\
\mathbf{x}_i, && \text{if } k = j \\
0, && \text{otherwise}
\end{cases}
$$
The equation is nothing but
$$
f = \sum_{j=1}^N a_{kj} \mathbf{x}_j + \sum_{i=1}^N a_{ik} \mathbf{x}_i
$$
Almost done! Now we only need some simplification. Recall the very simple rule that
$$
\sum_{i=1}^N x_i y_i = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}^T \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix} =  \mathbf{x}^T \mathbf{y}
$$
Thus
$$
\begin{align}
f &= \text{(k-th row of A) } \mathbf{x} + \text{(k-th column of A)}^T \mathbf{x}
\end{align}
$$
Now it is time to compute the gradient from partial derivative!
$$
\begin{align}
\nabla_\mathbf{x} \mathbf{x}^T A \mathbf{x}
& = \begin{bmatrix} 
\frac{\partial \mathbf{x}^T A \mathbf{x}}{\partial x_1} \\ \vdots \\ \frac{\partial \mathbf{x}^T A \mathbf{x}}{\partial x_k} \\ \vdots \\ \frac{\partial \mathbf{x}^T A \mathbf{x}}{\partial x_N} \\
\end{bmatrix}
= \begin{bmatrix} 
\vdots \\ \text{(k-th row of A) } \mathbf{x} + \text{(k-th column of A)}^T \mathbf{x} \\ \vdots
\end{bmatrix} \\
&= \left( \begin{bmatrix} \vdots \\ \text{(k-th row of A) } \\ \vdots \end{bmatrix} + \begin{bmatrix} \vdots \\ \text{(k-th column of A) }^T \\ \vdots \end{bmatrix} \right) \mathbf{x} \\
&= (A + A^T)\mathbf{x}
\end{align}
$$
So we are done!! The answer is:
$$
\nabla_\mathbf{x} \mathbf{x}^T A \mathbf{x} = (A + A^T)\mathbf{x}
$$
A: The only thing you need to remember/know is that
$$\dfrac{\partial (x^Ty)}{\partial x} = y$$
and the chain rule, which goes as
$$\dfrac{d(f(x,y))}{d x} = \dfrac{\partial (f(x,y))}{\partial x} + \dfrac{d( y^T(x))}{d x} \dfrac{\partial (f(x,y))}{\partial y}$$
Hence,
$$\dfrac{d(b^Tx)}{d x} = \dfrac{d (x^Tb)}{d x} = b$$
$$\dfrac{d (x^TAx)}{d x} = \dfrac{\partial (x^Ty)}{\partial x} +  \dfrac{d (y(x)^T)}{d x} \dfrac{\partial (x^Ty)}{\partial y}$$
where $y = Ax$. And then, that is,
$$\dfrac{d (x^TAx)}{d x} = \dfrac{\partial (x^Ty)}{\partial x} +  \dfrac{d( y(x)^T)}{d x} \dfrac{\partial (x^Ty)}{\partial y} = y + \dfrac{d (x^TA^T)}{d x} x = y + A^Tx = (A+A^T)x$$
A: Yet another approach.
We will utilize the following the identities 


*

*Trace and Frobenius product relation $$\left\langle A, B \right\rangle={\rm tr}(A^TB) = A:B$$ or $$\left\langle A^T, B \right\rangle ={\rm tr}(AB) = A^T:B$$

*Cyclic property of Trace/Frobenius product 
\begin{align}
\left\langle A, B C \right\rangle \equiv A:BC 
 &= AC^T:B\\
 &= B^TA:C\\
 &= BC : A\\
 &= {\text{etc.}} \cr
\end{align}
Let $f(x) := \left( \frac{1}{2} x^T A x - b^T x + c \right) $.
We obtain the differential first, and then the gradient subsequently.
\begin{align}
 d\,f(x) 
&=  d\left( \frac{1}{2} x^T A x - b^T x + c \right) \\
&=  d\left( \frac{1}{2} \left( x: A x \right) - \left( b : x \right) + c \right) \\
&=  \frac{1}{2} \left[ \left( dx: A x \right) + \left( x: A dx \right) \right] - \left( b : dx \right)  \\
&=  \frac{1}{2} \left[ \left( A x : dx \right) + \left( A^Tx: dx \right) \right] - \left( b : dx \right)  \\
&=  \frac{1}{2}  \left[ \left(A + A^T  \right)x: dx \right] - \left( b : dx \right)  \\
&=  \left( \frac{1}{2}  \left[ \left(A + A^T  \right)x\right] - b \right): dx \\
\end{align}
Thus the gradient is 
$$ \eqalign {
 \frac { \partial} {\partial x}f(x) &=  \frac{1}{2}  \left[ \left(A + A^T  \right)x\right] - b \\ &=   \frac{1}{2} A^T x +  \frac{1}{2} A x - b . \cr
} $$
A: Let $f : \mathbb R^n \to \mathbb R$ be defined by
$$f (\mathrm x) := \rm x^\top A \, x$$
Hence,
$$f (\mathrm x + h \mathrm v) = (\mathrm x + h \mathrm v)^\top \mathrm A \, (\mathrm x + h \mathrm v) = f (\mathrm x) + h \, \mathrm v^\top \mathrm A \,\mathrm x + h \, \mathrm x^\top \mathrm A \,\mathrm v + h^2 \, \mathrm v^\top \mathrm A \,\mathrm v$$
Thus, the directional derivative of $f$ in the direction of $\rm v$ at $\rm x$ is
$$\lim_{h \to 0} \frac{f (\mathrm x + h \mathrm v) - f (\mathrm x)}{h} = \mathrm v^\top \mathrm A \,\mathrm x + \mathrm x^\top \mathrm A \,\mathrm v = \langle \mathrm v , \mathrm A \,\mathrm x \rangle + \langle \mathrm A^\top \mathrm x , \mathrm v \rangle = \left\langle \mathrm v , \color{blue}{\left(\mathrm A + \mathrm A^\top\right) \,\mathrm x} \right\rangle$$
Lastly, the gradient of $f$ with respect to $\rm x$ is
$$\nabla_{\mathrm x} \, f (\mathrm x) = \color{blue}{\left(\mathrm A + \mathrm A^\top\right) \,\mathrm x}$$

multivariable-calculus scalar-fields gradient quadratic-forms
A: I am just writing this answer for future reference and for clarity because the accepted answer is not completely correct and may case confusion.
I will use a simple proof to better understand the multiplication rule in the calculation of $\nabla\mathbf{x^TAx}$.
For $\mathbf{x}\in \mathbb{R}^n$ and $\mathbf{A}\in \mathbb{R}^{n\times n}$ let: $$f(g(\mathbf{x}),h(\mathbf{x}))=\langle g(\mathbf{x}),h(\mathbf{x})\rangle=g^T(\mathbf{x})h(\mathbf{x})$$
Where:
\begin{equation}
\begin{split}
&g(\mathbf{x})=\mathbf{x}\\
&h(\mathbf{x})=\mathbf{Ax}
\end{split}
\end{equation}
From the definition of $f$, it is obvious that $f(g(\mathbf{x}),h(\mathbf{x}))=\mathbf{x^TAx}$.
In order to calculate the derivative, we will use the following fundamental properties, where $\mathbf{I}$ is the identity matrix:
\begin{equation}
\begin{split}
&\dfrac{\partial \mathbf{A^Tx}}{\partial\mathbf{x}}=\dfrac{\partial \mathbf{x^TA}}{\partial\mathbf{x}}=\mathbf{A^T}\\
&\dfrac{\partial \mathbf{x}}{\partial\mathbf{x}}=\mathbf{I}
\end{split}
\end{equation}
Hence, from the multiplication rule (you can see the rule at wikipedia), we got:
\begin{equation}
\begin{split}
\dfrac{df(g(\mathbf{x}),h(\mathbf{x}))}{d\mathbf{x}}&=g^T(\mathbf{x})\dfrac{\partial h(\mathbf{x})}{\partial\mathbf{x}}+h^T(\mathbf{x})\dfrac{\partial g(\mathbf{x})}{\partial\mathbf{x}}=\\
&=\mathbf{x^T}\dfrac{\partial \mathbf{Ax}}{\partial\mathbf{x}}+(\mathbf{Ax})^T\dfrac{\partial \mathbf{x}}{\partial\mathbf{x}}=\\
&=\mathbf{x^TA}+\mathbf{x^TA^TI}=\\
&=\mathbf{x^TA}+\mathbf{x^TA^T}=\\
&=\mathbf{x^T}(\mathbf{A+A^T})
\end{split}
\end{equation}
As a result, from the definition of gradient, we got:
$$
\nabla f=\Bigg(\dfrac{df}{d\mathbf{x}}\Bigg)^T=(\mathbf{x^T}(\mathbf{A+A^T}))^T=(\mathbf{A^T+A})\mathbf{x}
$$
Note: The reason I did the proof this way is to be more generalizable. So you can can plug arbitrary functions $g$ and $h$ and use the above multiplication rule to derive the result.
