Okay so I'm really stuck on this problem, have stared at it for a few hours and am quite desperate now. Does anyone have an idea?

Let A=($\displaystyle a_{ij}$) be a symmetric nxn matrix, i.e. $\displaystyle a_{ij}=a_{ji} $ for all i,j in {i,...,n}. Define $\displaystyle f: R^{n} \rightarrow R $ by $\displaystyle f(\vec{x}) = \frac {1} {2} \vec{x}^{T} A \vec{x} $ , where the superscript T denotes transpose so that $\displaystyle f( x_{1} , ... , x_{n} ) = \frac {1} {2} \sum\limits_{i,j=1}^{n} a_{ij} x_{i} x_{j} $ . Show that $\displaystyle (\nabla f) (\vec{x}) = A (\vec{x}) $ .