I'm trying to differentiate the following..

$\displaystyle \frac{dG}{dz} = \frac{1-a}{1+a} * \frac{2a}{(1-az)^{2}}$

z is evaluated at 1 according to the Leibniz notation.

The answer is $\displaystyle \frac{2a}{1-a^2}$

I'm not quite sure how to get there.

$\displaystyle (1-az)^{-2} $ would be differentiated with the chain rule to give..

$\displaystyle -2(1-az)^{-3} * (-a) = \frac{2a}{(1-az)^{3}} $

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.