Chapter 10: Infinite Sequences and Series - Section 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 634: 61

$1-2x+3x^2-4x^3+.....; |x| \lt 1$

Recall the Taylor series for $\dfrac{-1}{1+x}=-\dfrac{-1}{1-(-x)}= -1+x-x^2+x^3-......+x^n+.....; |x| \lt 1$ Now, $\dfrac{d}{dx}(\dfrac{-1}{1+x}) =\dfrac{d}{dx} [ -1+x-x^2+x^3-......+x^n+.....]$ or, $\dfrac{d}{dx}(\dfrac{-1}{1+x})=\dfrac{1}{(1+x)^2} =0+1-2x+3x^2-4x^3+.....$ or, $\dfrac{d}{dx}(\dfrac{-1}{1+x})= \dfrac{1}{(1+x)^2} =1-2x+3x^2-4x^3+.....; |x| \lt 1$