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