What is maclaurin e^x(1+x)^{-1} ?

The answer to maclaurin e^x(1+x)^{-1} is 1+1/2 x^2-1/3 x^3+3/8 x^4-11/30 x^5+\ldots

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maclaurin e^x(1+x)^{-1}

maclaurin

e^{x} (1 + x)^{- 1}

1 + \frac{1}{2} x^{2} - \frac{1}{3} x^{3} + \frac{3}{8} x^{4} - \frac{11}{30} x^{5} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{1}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{9}{4 !} x^{4} + \frac{- 44}{5 !} x^{5} + \dots

Refine

= 1 + \frac{1}{2} x^{2} - \frac{1}{3} x^{3} + \frac{3}{8} x^{4} - \frac{11}{30} x^{5} + \dots

\frac{\partial}{\partial x} (2 x + 2 y)

f (x) = \frac{1}{x - 3}

\int \frac{( x + 6 )}{x ^{2} + 9} d x

\frac{s + 1}{s ^{2} - 3 s + 2}

\frac{\partial}{\partial x} (9 x + y + 7 x y)

What is maclaurin e^x(1+x)^{-1} ?
The answer to maclaurin e^x(1+x)^{-1} is 1+1/2 x^2-1/3 x^3+3/8 x^4-11/30 x^5+\ldots