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

The answer to maclaurin x/(1+4x^2) is x-4x^3+16x^5-64x^7+256x^9+\ldots

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maclaurin x/(1+4x^2)

maclaurin

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

x - 4 x^{3} + 16 x^{5} - 64 x^{7} + 256 x^{9} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 24}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{1920}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 322560}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{92897280}{9 !} x^{9} + \dots

Refine

= x - 4 x^{3} + 16 x^{5} - 64 x^{7} + 256 x^{9} + \dots

\frac{d}{d x} (e^{x} sin (x))

y = sin (9 x) + sin^{2} (9 x), (0, 0)

\frac{d y}{d x} + x^{2} sec (y) = 0, y (0) = 1

\frac{d}{d x} (3 - 8 x)

\int \frac{2 x ^{2} + x + 3}{( x + 1 ) ( x ^{2} + 1 )} d x

What is maclaurin x/(1+4x^2) ?
The answer to maclaurin x/(1+4x^2) is x-4x^3+16x^5-64x^7+256x^9+\ldots