What is taylor x/((1+4x^2)^2) ?

The answer to taylor x/((1+4x^2)^2) is x-8x^3+48x^5-256x^7+1280x^9+\ldots

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taylor x/((1+4x^2)^2)

taylor

\frac{x}{( 1 + 4 x ^{2} ) ^{2}}

x - 8 x^{3} + 48 x^{5} - 256 x^{7} + 1280 x^{9} + \dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( \frac{x}{( 1 + 4 x ^{2} ) ^{2}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{x}{( 1 + 4 x ^{2} ) ^{2}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{x}{( 1 + 4 x ^{2} ) ^{2}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 48}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{5760}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 1290240}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{464486400}{9 !} x^{9} + \dots

Refine

= x - 8 x^{3} + 48 x^{5} - 256 x^{7} + 1280 x^{9} + \dots

\frac{\partial}{\partial z} (x^{2} y + z)

y^{^{''}} + 10 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 0

(x sin (x))^{^{'}}

\frac{9}{s ^{2} + 9 s + 9}

(e^{- ∣ x ∣})^{^{'}}

What is taylor x/((1+4x^2)^2) ?
The answer to taylor x/((1+4x^2)^2) is x-8x^3+48x^5-256x^7+1280x^9+\ldots