What is maclaurin (1+x)^{1/6} ?

The answer to maclaurin (1+x)^{1/6} is 1+1/6 x-5/72 x^2+55/1296 x^3-935/31104 x^4+\ldots

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maclaurin (1+x)^{1/6}

maclaurin

(1 + x)^{\frac{1}{6}}

1 + \frac{1}{6} x - \frac{5}{72} x^{2} + \frac{55}{1296} x^{3} - \frac{935}{31104} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( ( 1 + x ) ^{\frac{1}{6}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ( 1 + x ) ^{\frac{1}{6}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ( 1 + x ) ^{\frac{1}{6}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{\frac{1}{6}}{1 !} x + \frac{- \frac{5}{36}}{2 !} x^{2} + \frac{\frac{55}{216}}{3 !} x^{3} + \frac{- \frac{935}{1296}}{4 !} x^{4} + \dots

Refine

= 1 + \frac{1}{6} x - \frac{5}{72} x^{2} + \frac{55}{1296} x^{3} - \frac{935}{31104} x^{4} + \dots

\int_{0}^{2 π} ∣ sin (x) ∣ d x

\frac{d}{d x} (\frac{x}{9 + x ^{2}})

f (x) = (6 x - x^{3}) (5 x + 2)

\int_{1}^{e^{5}} x^{3} ln (x) d x

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

What is maclaurin (1+x)^{1/6} ?
The answer to maclaurin (1+x)^{1/6} is 1+1/6 x-5/72 x^2+55/1296 x^3-935/31104 x^4+\ldots