Derivatives problem

**notgoodatmath** · June 7th 2010, 01:18 AM

I'm trying to differentiate the following..

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

z is evaluated at 1 according to the Leibniz notation.

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

I'm not quite sure how to get there.

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

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

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.

**p0oint** · June 7th 2010, 02:19 AM

Its already differentiated, can't you see?

dG/dz means that you already differendiated it. Now just substitute for z=1.

Now you got:

$\frac{dG}{dz} |_{z=1}= \frac{1-a}{1+a} * \frac{2a}{(1-a)^{2}}=\frac{1-a}{1+a}*\frac{2a}{(1-a)*(1-a)}=\frac{2a}{(1-a)*(1+a)}$

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