What is maclaurin (1+x^2)^{(-3)/2} ?

The answer to maclaurin (1+x^2)^{(-3)/2} is 1-3/2 x^2+15/8 x^4-35/16 x^6+315/128 x^8+\ldots

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maclaurin (1+x^2)^{(-3)/2}

maclaurin

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

1 - \frac{3}{2} x^{2} + \frac{15}{8} x^{4} - \frac{35}{16} x^{6} + \frac{315}{128} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 3}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{45}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 1575}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{99225}{8 !} x^{8} + \dots

Refine

= 1 - \frac{3}{2} x^{2} + \frac{15}{8} x^{4} - \frac{35}{16} x^{6} + \frac{315}{128} x^{8} + \dots

y^{^{'}} = \frac{( 3 y ^{2} + 9 )}{2 x y}

n = 1 \sum \infty \frac{1}{2 + n}

x \to 0 lim (\frac{1 - e ^{2 x}}{x})

\int_{- \infty}^{8} r e^{\frac{r}{2}} d r

\frac{d}{d t} (sin (π t))

What is maclaurin (1+x^2)^{(-3)/2} ?
The answer to maclaurin (1+x^2)^{(-3)/2} is 1-3/2 x^2+15/8 x^4-35/16 x^6+315/128 x^8+\ldots