# w=f(x-y,y-z,z-x) show that dw/dx+dw/dy+dwdz=0

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f is a function of 3 variables.

Let `(del f)/(del x_1) ` be the derivative of f with respect to the first variable.

Let be the derivative of f with respect to the second variable.

Let `(del f)/(del x_3)` be the derivative of f with respect to the third variable.

`(dw)/(dx)=(delf)/(delx_1) (x-y,y-z,z-x)*(del(x-y))/(delx)+(delf)/(del x_2) (x-y,y-z,z-x)*(del(y-z))/(del x)+(delf)/(delx_3) (x-y,y-z,z-x)*(del(z-x))/(delx)`

`(dw)/(dx)=(delf)/(delx_1) (x-y,y-z,z-x)+0-(delf)/(delx_3) (x-y,y-z,z-x)`

`(dw)/(dy)=(delf)/(delx_1) (x-y,y-z,z-x)*(del(x-y))/(dely)+(delf)/(del x_2) (x-y,y-z,z-x)*(del(y-z))/(del y)+(delf)/(delx_3) (x-y,y-z,z-x)*(del(z-x))/(dely)`

`(dw)/(dy)=-(delf)/(delx_1) (x-y,y-z,z-x)+(delf)/(del x_2) (x-y,y-z,z-x)+0`

`(dw)/(dz)=(delf)/(delx_1) (x-y,y-z,z-x)*(del(x-y))/(delz)+(delf)/(del x_2) (x-y,y-z,z-x)*(del(y-z))/(del z)+(delf)/(delx_3) (x-y,y-z,z-x)*(del(z-x))/(delz)`

` <br> `

`(dw)/(dz)=0-(delf)/(del x_2) (x-y,y-z,z-x)+(delf)/(delx_3) (x-y,y-z,z-x)`

`(dw)/(dx)+(dw)/(dy)+(dw)/(dz)=(delf)/(delx_1) (x-y,y-z,z-x)-(delf)/(del x_3) (x-y,y-z,z-x)-(delf)/(delx_1) (x-y,y-z,z-x)+(delf)/(del x_2) (x-y,y-z,z-x)-(delf)/(del x_2) (x-y,y-z,z-x)+(delf)/(delx_3) (x-y,y-z,z-x)`

Therefore

`(dw)/(dx)+(dw)/(dy)+(dw)/(dz)=0`