What is maclaurin (1+x)^{-4/3} ?

The answer to maclaurin (1+x)^{-4/3} is 1-4/3 x+14/9 x^2-140/81 x^3+455/243 x^4+\ldots

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maclaurin (1+x)^{-4/3}

maclaurin

(1 + x)^{- \frac{4}{3}}

1 - \frac{4}{3} x + \frac{14}{9} x^{2} - \frac{140}{81} x^{3} + \frac{455}{243} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- \frac{4}{3}}{1 !} x + \frac{\frac{28}{9}}{2 !} x^{2} + \frac{- \frac{280}{27}}{3 !} x^{3} + \frac{\frac{3640}{81}}{4 !} x^{4} + \dots

Refine

= 1 - \frac{4}{3} x + \frac{14}{9} x^{2} - \frac{140}{81} x^{3} + \frac{455}{243} x^{4} + \dots

\frac{\partial}{\partial x} (\frac{1}{ln ( \frac{y}{x} )})

\frac{s ^{2}}{( s ^{2} + 1 ) ^{2}}

ln (\frac{( x + 6 ) ^{4} e ^{5 x^{3} - 3}}{x + 2 x ^{3}})

y = x^{2} + 4 x + 7, at x = 1

\frac{7}{2 7 x + 1}, at x = 9

What is maclaurin (1+x)^{-4/3} ?
The answer to maclaurin (1+x)^{-4/3} is 1-4/3 x+14/9 x^2-140/81 x^3+455/243 x^4+\ldots