How is the mixed 2nd partial derivative simplified to a more efficient form?

I'm implementing the Finite Different 2nd and 3rd derivatives in my research and naturally I'm looking for the most efficient approach.

From https://en.wikipedia.org/wiki/Finite_difference#cite_ref-WilmottHowison1995_1-0 I found a more efficient formula that takes 2 function evaluations out of the picture. But I'm not sure what simplification or substitution was implemented. Can anyone help me out? See the Image below.

Thank you!

• Try to move this question in the mathematics community. – james watt Sep 24 '20 at 12:05
• @JamesWatt: I think the question is completely appropriate here. – davidhigh Sep 24 '20 at 15:34

The first formula evaluates the derivative on the points $$\Big((x-h,y-k), (x-h,y+k), (x+h,y-k),(x+h,y+k)\Big)$$, the second formula uses another stencil that involves points that are also used by the first derivative.