Answer
$\dfrac{1}{(1+x)^2} =1-2x+3x^2-4x^3+.....; |x| \lt 1$
Work Step by Step
The Taylor series for $\dfrac{-1}{1+x} $ can be defined as: $\dfrac{-1}{1+x}=-\dfrac{-1}{1-(-x)}= -1+x-x^2+x^3-......+x^n+.....; |x| \lt 1$
Consider the given series:
$\dfrac{d}{dx}(\dfrac{-1}{1+x}) =\dfrac{d}{dx} [ -1+x-x^2+x^3-......+x^n+.....]$
or, $\dfrac{1}{(1+x)^2} =0+1-2x+3x^2-4x^3+.....$
or, $\dfrac{1}{(1+x)^2} =1-2x+3x^2-4x^3+.....; |x| \lt 1$
