What's gone wrong in my example to show that $\frac{\partial}{\partial \underline c}(A\underline c) = A$? First, an example:
Suppose $A$ is a symmetric matrix 
$A = \begin{bmatrix}
a & b \\
b & a \\
\end{bmatrix}$, 
and 
$\underline c = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$
Then 
$A \underline c = 
\begin{bmatrix}
ac_1 + b c_2 \\
bc_1 + ac_2 \\
\end{bmatrix}$
and hence, $\frac{\partial}{\partial \underline c}(A\underline c) = 
\begin{bmatrix}
\frac{\partial}{\partial c_1}(ac_1 + b c_2) \\
\frac{\partial}{\partial c_2}(bc_1 + ac_2) \\
\end{bmatrix} = 
\begin{bmatrix}
a
\\a
\end{bmatrix} \neq A$ 
However, I'm told that the derivative with respect to the vector of the product of a symmetric $(n\times n)$ matrix $A$ and an $(n\times 1)$ vector $\underline c$ is equal to the matrix $A$, such that,
$\frac{\partial}{\partial \underline c}(A\underline c) = A$
Where is the error in my understanding?
 A: You have a vector result $$ y = Ac$$
Thus, you are  computing a vector-by-vector gradient which results in a second-order tensor, ie. a matrix, with the following form:
$$ \frac{\partial y}{\partial c}=
\begin{bmatrix}
{\frac {\partial y_{1}}{\partial c_{1}}}&{\frac {\partial y_{1}}{\partial c_{2}}}\\
{\frac {\partial y_{2}}{\partial c_{1}}}&{\frac {\partial y_{2}}{\partial c_{2}}}\\
\end{bmatrix}
$$
You are just missing the off-diagonal terms.
A: Since your function is vector valued, its derivative is a matrix : $\left(\frac{\partial \mathbf{u}}{\partial\mathbf{x}}\right)_{i,j} = \frac{\partial u_i}{\partial x_j}$.
Hence, $\frac{\partial A \mathbf{c}}{\partial \mathbf{c}} = \begin{bmatrix} \frac{\partial a c_1 + b c_2}{\partial c_1} & \frac{\partial a c_1 + b c_2}{\partial c_2} \\ \frac{\partial b c_1 + a c_2}{\partial c_1} & \frac{\partial b c_1 + a c_2}{\partial c_2}\end{bmatrix} = \begin{bmatrix} a & b\\ b & a\end{bmatrix}$.
Remark : in general, $\frac{\partial A \mathbf{c}}{\partial \mathbf{c}} = A^T $, but here your matrix is symmetric (i.e. $A=A^T$).
A: Your definition of the derivative
$
\frac{\partial}{\partial \overline{c}}.
$
The function $$\overline{c} \mapsto A \overline{c}$$ is total differentiable if there exists a linear map $B$, with
$$ 
\lim_{\|h\| \rightarrow 0} \frac{\|A\left(\overline{c}+\overline{h}\right) - A\overline{c}  - B\overline{h}\|}{\|\overline{h}\|} = 0
$$. 
Since the map is linear, it is obvious that $A = B$, so the total derivative is $A$.
