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Math Help - Derivatives problem

  1. #1
    Newbie
    Joined
    Oct 2008
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    17

    Derivatives problem

    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.
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  2. #2
    Member
    Joined
    Nov 2009
    Posts
    130
    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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