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

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

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

maclaurin

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

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

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{3}{2}}{1 !} x + \frac{\frac{3}{4}}{2 !} x^{2} + \frac{- \frac{3}{8}}{3 !} x^{3} + \frac{\frac{9}{16}}{4 !} x^{4} + \dots

Refine

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

\frac{d}{d x} (1 0^{1 - x^{2}})

\frac{\partial}{\partial x} (arctan (x + 4 y))

x \to - 2 lim (\frac{x ^{2} + 8}{x - 2})

y^{^{''}} = - 3 y - 4 y^{^{'}}

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