I see that the first derivative of variance, wrt $x_i$ is $\frac{2}{N}(x_i - \mu)$. And the second derivative would be simply $\frac{2}{N}$ (beyond this, all subsequent derivatives should be zero.)
So Should the Taylor Series Expansion simply include these two terms?
Generally if you take the Taylor series of a multivariate function you will invoke all of the partial derivatives, including the cross-term ones (e.g. $\frac{\partial^2}{\partial x_i \partial x_j}$).
However, broadly you are correct - the variance is purely quadratic in the $x_i$ terms, so if you take its Taylor series you will get back the exact same quadratic polynomial. And because the quadratic doesn't actually include any cross-multiplied terms, the second derivatives $\frac{\partial^2}{\partial x_i \partial x_j}$ only exist when $i = j$ anyway.
