What is maclaurin 1/((1-x^2)^2) ?

The answer to maclaurin 1/((1-x^2)^2) is 1+2x^2+3x^4+4x^6+5x^8+\ldots

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maclaurin 1/((1-x^2)^2)

maclaurin

\frac{1}{( 1 - x ^{2} ) ^{2}}

1 + 2 x^{2} + 3 x^{4} + 4 x^{6} + 5 x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( \frac{1}{( 1 - x ^{2} ) ^{2}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{( 1 - x ^{2} ) ^{2}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{( 1 - x ^{2} ) ^{2}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{4}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{72}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{2880}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{201600}{8 !} x^{8} + \dots

Refine

= 1 + 2 x^{2} + 3 x^{4} + 4 x^{6} + 5 x^{8} + \dots

10 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0

\frac{d r}{d θ} = \frac{r ^{2} + 5 r θ ^{4}}{θ ^{5}}

y = \frac{2 x}{x ^{2} - 7}, (3, 3)

x, 2 x 1 - x^{2}

x \to 2 lim (x (x - 1) (x + 1))

What is maclaurin 1/((1-x^2)^2) ?
The answer to maclaurin 1/((1-x^2)^2) is 1+2x^2+3x^4+4x^6+5x^8+\ldots