Let arbitrary function f(x,y)=ax2+bxy+cy2; x=u+log(v), y=u-log(v).
f(u,v)=a(u2+2ulog(v)+log2(v))+b(u2-log2v)+c(u2-2ulog(v)+log2(v))
∂y/∂u=∂x/∂u=1. So ∂f/∂u=(∂f/∂x)(∂x/∂u)=(∂f/∂y)(∂y/∂u), making ∂f/∂u=∂f/∂x=∂f/∂y. This means that u is manipulated equivalently to x and y by any function and its derivatives. Also, since both x and y involve u, u combines the coefficients of x and y. In the example, for example, the coefficients of the squares of x and y, that is, a and c, are combined into a+c for the square of u.
Since ∂2f/∂x∂y=∂2f/∂y∂x then ∂2f/∂x∂y+∂2f/∂y∂x=2∂2f/∂x∂y=2∂2f/∂y∂x. u is effectively the joint expression of x and y. v is not involved in any of the differentials. If it were this equivalence would not be valid because ∂x/∂v≠∂y/∂v, and neither equals 1.
∂f/∂x=2ax+by, ∂f/∂y=bx+2cy, ∂f/∂u=2au+2alog(v)+2bu+2uc-2clog(v).
∂2f/∂x2=2a, ∂2f/∂y2=2c, ∂2f/∂x∂y=b=∂2f/∂y∂x.
∂2f/∂u2=2a+2b+2c.
∂2f/∂x2+2∂2f/∂x∂y+∂2f/∂y2=2a+2b+2c.
So ∂2f/∂u2=∂2f/∂x2+2∂2f/∂x∂y+∂2f/∂y2.