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

The answer to maclaurin (1+x^2)^{3/2} is 1+3/2 x^2+3/8 x^4-1/16 x^6+3/128 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{3}{8} x^{4} - \frac{1}{16} x^{6} + \frac{3}{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{9}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 45}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{945}{8 !} x^{8} + \dots

Refine

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

\frac{d}{d x} (\frac{6 x}{( x ^{2} - 1 ) ^{2}})

\frac{\partial}{\partial x} ((3 x y - 2)^{2})

\int_{4}^{9} x d x

f (x) = sin (x) + x cos (x)

f (x) = x + 3

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