Evaluation of this complex derivative $\frac{d}{dz}(i\frac{z+i}{z-i})$
I wrote $\frac{d}{dz} = \frac{1}{2}(\frac{d}{dx}-i\frac{d}{dy})$  and expand $z$ into $x + iy$. But at the end or in the middle i cannot group $x+iy$ term again completely so the answer cannot be expressed in $z$. 
How to evaluate this complex derivative so the answer can be expressed in $z$?
Thanks in advance.

if you write $f(z)=i\frac{z+i}{z-i} = \frac{u}{v}$ then:
$$
u=-y-1+xi \\
v=x+iy-i
$$
using subscripts for partial differentials:
$$
2\frac{df}{dz}=v^{-2}(u_x v-uv_x -i(u_yv-uv_y))
$$
here the term $v^{-2}$ is already a function of $z$ so we check:
$$
u_x v-uv_x = i(x+iy-i) - (-y-1+xi) = 2\\
u_yv-uv_y = -1(x+iy-i) - (-y-1+xi)i=2i
$$
from which:
$$
u_x v-uv_x -i(u_yv-uv_y) = 2 -i2i = 4
$$
so finally
$$
\frac{df}{dz}=\frac2{(z-i)^2}
$$
which can be checked by ordinary differentiation wrt the complex variable $z$
