What's the partial derivative of this expression?

Question

How do I show that $\dfrac{\partial{x'Ax}}{\partial{x_i}} = 2(Ax)_i$, where $x$ is an $n\times1$ column vector and $A$ is a $n \times n$ symmetric matrix. Any lucid and memorable proof?

Answer 1

This is how i remember this by using the total derivative of the expression:

$$d(x'Ax)=dx'Ax+x'Adx$$.

The expressions are scalars (easy to check by format inspection) so transposing the second expression will not change it. Also remember that transposing changes the order of terms.

$$d(x'Ax)=dx'Ax+dx'A'x=dx'(A+A')x$$

As A is symmetric (A'=A) we obtain:

$$d(x'Ax)=dx'(2Ax)$$.

Now remember that the total derivative can be written as the dot product of the gradient and the infinitesimal vector dx. Hence, the expression $2Ax$ is the gradient (what you were looking for).