Poisson relationship between area and length of electric wire

Question

I was given the following formula to relate the change of area against the change of length of an electric wire with a Poisson ratio:
$ {Delta A over A} = -2 nu {Delta L over L} $
where $ Delta A over A $ represents the change in cross-sectional area of the wire due to the transverse strain as the wire gets pulled longitudinally stretching length $ L $ to $ L+Delta L$.
I don't get how this equation is derived. The Poisson ratio is defined by $ nu = -{epsilon_{lateral} over {epsilon_{longitudinal}} }  = - {{Delta d / d} over {Delta L / L}} $ where $ d $ is the diameter of the cross section. Then the ratio of the area:
$ {Delta A over A} = {{0.25pi(d+Delta d)^2 - 0.25pi d^2} over {0.25pi d^2}} = {{2d Delta d} over d^2} + {{Delta d^2} over {d^2}} = -2 nu { Delta L over L} + big( nu {Delta L over L} big)^2 neq -2 nu {Delta L over L} $

NMech · Accepted Answer

Since you are essentially using infinitesimal changes, then higher order differences can be neglected.
I.e. following from your equation
$${Delta A over A} = {{0.25pi(d+Delta d)^2 - 0.25pi d^2} over {0.25pi d^2}} = {{2d Delta d} over d^2} + {{Delta d^2} over {d^2}} = -2 nu { Delta L over L} + big( nu {Delta L over L} big)^2 $$
because $left(frac{Delta L}{L}right)^2$ is a second order difference, you can assume that $left( nu {Delta L over L} right)^2approx 0$.
Therefore:
$$  -2 nu { Delta L over L} + underbrace{left( nu {Delta L over L} right)^2}_{approx 0}approx -2 nu { Delta L over L} $$

Poisson relationship between area and length of electric wire

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