Chapter 10: Infinite Sequences and Series - Section 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 634: 62

$$2x+4x^3+6x^5.....$$

Recall the Taylor series for $\dfrac{1}{1-x^2} $ can be defined as: $\dfrac{1}{1-x^2}=1+x^2+x^4-......+x^n+.....; |x| \lt 1$ Now, $\dfrac{d}{dx}(\dfrac{1}{1-x^2}) =\dfrac{d}{dx} [1+x^2+x^4+x^6......+x^n+.....]$ or, $\dfrac{d}{dx}(\dfrac{1}{1-x^2}) =\dfrac{2x}{(1-x^2)^2} =0+2x+4x^3+6x^5.....$ or, $\dfrac{d}{dx}(\dfrac{1}{1-x^2}) =\dfrac{2x}{(1-x^2)^2} =2x+4x^3+6x^5.....$