Its already differentiated, can't you see?
dG/dz means that you already differendiated it. Now just substitute for z=1.
Now you got:
I'm trying to differentiate the following..
z is evaluated at 1 according to the Leibniz notation.
The answer is
I'm not quite sure how to get there.
would be differentiated with the chain rule to give..
Does the z disappear because we substitute in the value 1 at this point?
I'm not sure how to get rid of the (1+a) in the denominator.