If you were to want the entire Taylor series, I would start by taking the necessary number of derivatives to make it to the #x^2# term (i.e. the #n = 2# term). That means up to and including the 2nd derivative.
#color(green)(f^((0))(x)) = f(x) = (1+x)^(-2) color(green)(= 1/(1+x)^2)#
#color(green)(f'(x)) = -2(1+x)^(-3) = color(green)(-2/(1+x)^3)#
#color(green)(f''(x)) = 6(1+x)^(-4) = color(green)(6/(1+x)^4)#
The general formula for the Taylor series is:
#sum_(n=0)^N f^((n))(a)/(n!)(x-a)^n#
For #N = 2# (otherwise known as "truncating at the second order term") and centered around #a = 0#, this is:
#= f^((0))(0)/(0!)(x-0)^0 + (f'(0))/(1!)(x-0)^1 + (f''(0))/(2!)(x-0)^2#
#= (1/(1+(0))^2)/(0!)*1 + (-2/(1+(0))^3)/(1!)x + (6/(1+(0))^4)/(2!)x^2#
#= (1/(1)^2) + (-2/(1)^3)x + (6/(2*(1)^4))x^2#
#= 1 - (2/1)x + (6/2)x^2#
#= color(blue)(1 - 2x + 3x^2)#
So your coefficient is #3#. Now, you could have also done only the 2nd derivative, but this shows you the pattern for the entire Taylor series. You could write out more terms:
#= 1 - 2x + 3x^2 - 4x^3 + 5x^4 - ...#
Note that for the series, #a# does not vary, but #n# does. Also, #(x-a)^n# is not affected by #x->a#.
