I have the following equation which is from a famous paper in economics:
$$ \sum_{i=1}^n x_i \, dx_i=\frac{1}{2} d\Big[\sum_{i=1}^n x_{i}^2\Big]=XH \, dX+\frac{1}{2}d\big[X^2H\big] $$ Can you tell me how to get to the second and the third equality sign.
We have that
$$ \sum_i^n x_i=X $$
and
$$ \sum_i^n \Big(\frac{x_i}{X} \Big)^2=H $$
I think the "d" $dx_i$ is the same as in a derivative.
The $\rm d$ operator is the derivative, in a sense. The product (Leibniz) rule applies. And $\rm d$ is additive. Also, chain rule holds. So ${\rm d}(x_i^2) = 2 x_i {\rm d}x_i$. This indeed gives $$\frac{1}{2}{\rm d}\left(\sum x_i^2\right) = \frac{1}{2}\sum d(x_i^2) = \frac{1}{2} \sum 2x_i {\rm d}x_i = \sum x_i {\rm d}x_i,$$ as we wanted.
For the second part, are you sure that $H$ and $X$ are as stated? Look: $$H = \sum \left(\frac{x_i}{X}\right)^2 = \frac{1}{X^2}\sum x_i^2 \implies X^2H = \sum x_i^2 \implies {\rm d}(X^2H) = 2\sum x_i{\rm d}x_i.$$ With that factor $1/2$, we already have that term in the first equality. The $XH{\rm d}X$ parcel is left over.